UC-NRLF 


B    3    112 


Library 


FINANCIAL  ENGINEERING 


FINANCIAL 

ENGINEERING 


A  TEXT  FOR 

CONSULTING,  MANAGING  AND  DESIGNING 
ENGINEERS  AND  FOR  STUDENTS 


BY 

O.  B.  GOLDMAN 

CONSULTING  ENGINEER 

PROFESSOR   OF  HEAT    ENGINEERING,   THE    OREGON    STATE    AGRICULTURAL 

COLLEGE,   HONORARY    MEMBER   LOCAL    87,   I.    U.    STEAM    AND 

OPERATING    ENGINEERS,   MEMBER  OREGON 

SOCIETY    OF    ENGINEERS,    ETC. 


NEW  YORK 

JOHN  WILEY  &   SONS,   INC. 

LONDON:  CHAPMAN  &  HALL,  LIMITED 

1920 


TAAT-3 


Engineering 
Library 


COPYRIGHT,  IQ20,  BY 
O.  B.  GOLDMAN 


THE-PLIMPTON-PRESS-NOBWOOD'MASS-U'S-A 


PREFACE 

AN  engineer  must  know  the  properties  of  all  material  with 
which  he  comes  in  contact  and  he  must  understand 
thoroughly  the  action  and  limitations  of  all  machines  and 
instruments  which  he  is  called  upon  to  use  or  test,  as  well  as 
the  proper  application  of  the  same.  More  than  this,  he  should 
know  how  to  translate  engineering  factors  into  dollars  and 
cents.  In  addition  he  must  know  how  to  install  a  service  not 
necessarily  at  the  highest  mechanical  or  electrical  efficiency, 
which  may  prove  and  often  does  prove  far  too  expensive,  but 
always  with  regard  to  the  highest  financial  efficiency  so  the 
resulting  service  will  be  rendered  with  the  least  effort,  in  the 
preparation  for  the  service  and  its  actual  rendition.  This  is 
Financial  Engineering. 

Technical  Engineering  must  be  supplemented  by  Financial 
Engineering  to  make  a  complete  and  harmonious  system. 
What  is  demanded  of  the  Financial  Engineer  is  a  solution  in 
terms  of  money,  the  standard  measure  of  commerce.  Every 
engineer  in  a  responsible  position  has  felt  this  and  likewise 
the  need  of  a  definite,  scientific  method  of  determining  the 
comparative  value  of  all  things  which  he  must  use  and  the 
value  of  systems  and  of  investments  in  general.  He  has  felt 
the  need  of  a  correct  method  of  determining  the  financial 
efficiency  of  undertakings,  not  merely  as  a  whole,  but  element 
by  element,  so  that  all  losses  might  be  discovered,  and  so  that 
the  size  and  design  for  best  economy  might  be  determined.  It 
was  because  of  this  demand  that  the  author  devoted  so  much 
of  his  time,  over  a  period  of  fourteen  years,  to  the  development 
of  Financial  Engineering. 

Financial  Engineering  does  not  invade  the  field  of  Economics. 
As  a  science  it  is  founded  on  facts  as  all  true  science  must  be 
and  each  fact  is  thoroughly  checked.  Financial  Engineering 

V 

^47192 


vi  PREFACE 

is  just  as  applicable  to  a  farm  as  to  a  railroad,  just  as  applicable 
to  a  store  as  to  a  power  system.  It  extends  engineering  over 
business  and  administrative  problems. 

This  book  is  written  primarily  for  the  practicing  engineer. 
All  mathematical  deductions  are  worked  out  in  detail,  leaving 
no  gaps  for  the  reader  to  bridge.  Many  examples  are  also 
fully  worked  out,  to  illustrate  the  practical  applications  of  the 
technique.  The  author  has  found  by  experience,  that  with  the 
aid  of  an  instructor,  students  can  master  the  subject  well. 
Like  the  practicing  engineer,  they  are  greatly  interested  in  it, 
and  seek  it  with  greater  avidity  than  any  other  course  within 
the  author's  experience. 

The  work  in  its  various  stages  of  development  has  been 
repeatedly  submitted  to  engineers,  in  articles,  lectures  and 
addresses,  so  that  it  has  had  the  benefit  of  their  criticisms  and 
suggestions.  The  Author's  obligation  extends  to  so  many 
that  he  is  unable  to  do  justice  to  all.  Especial  obligation  is 
acknowledged  to  Messrs.  Kremers,  Johnson  and  Byrne,  and  to 
Professor  Teeter;  also  to  the  General  Electric  Co.,  the  Westing- 
house  Co.,  and  Gordon  and  Finkheimer,  of  Portland,  Oregon, 
for  authentic  data  on  the  performance  and  costs  of  engineering 
equipment  beyond  that  which  had  been  accumulated  by  the 
author. 

O.  B.  GOLDMAN. 


CONTENTS 

ARTICLE  CHAPTER  I.      INTRODUCTION  PAGE 

1.  Duties  of  the  Financial  Engineer i 

2.  Cost  Segregation  and  Cost  Analysis i 

3-4.     Profit 2 

5.     Value 3 

6-9.     Basis  of  Rates 3 

10.  Division  of  a  Service 6 

1 1 .  Fixed  Charges  and  Operating  Costs 6 

12-13.     Individual  and  Integral  Undertakings 7 

14.     Competition,  Law  of  Supply  and  Demand. ...  10 
15-18.     Replacement  Costs,  Market  Value  of  Outstand- 
ing Liabilities,  Capitalization  of  Earnings  . .  n 

CHAPTER  II.    COST  SEGREGATION 

19-20.     Interest  and  Rents 15 

21.  Depreciation  and  Appreciation 16 

22.  Obsolescence 17 

23.  Inadequacy 17 

24.  Uselessness 18 

25-26.     Natural  and  Operating  Life 18 

27.  Materials  Consumed 21 

28.  Attendance .- 22 

29.  Maintenance 22 

30.  Stock  Account 23 

31-32.     Storage,  Sale,  and  Distribution 23 

CHAPTER  III.    FUNDAMENTAL  FINANCIAL  CALCULATIONS 

33-34.     Simple  and  Compound  Interest 25 

35.     Principal 26 

vii 


Vlll 


CONTENTS 


36-39.     Equity N 27 

40-41.     Term  Factor 31 

42-47.     Depreciation  Rate 36 

48.  Present  Worth  of  a  Depreciating  Equipment. .  42 

49.  Vestance 44 

50.  Depreciation  Vestance 45 

5 1 .  Operating  Vestance 48 

52.  Total  Vestance 49 

53.  Annual  Operating  Cost 52 

54.  Taxes  and  Insurance 53 

CHAPTER  IV.    BASIC  COSTS 

55.  Basic  Costs 59 

56-57.     Records 60 

Prices  and  Operating  Costs  of: 

58-63.            Steam  Engines 62 

64-65.            Boilers 75 

66-67.            Buildings 78 

68-7 1 .            Centrifugal  Pumps 80 

72-77.            Motors  and  Generators 92 

78.            Oil  Engines 99 

78.  Gas  Producer  Engines 99 

79.  Diesel  Engines 101 

80-83.            Illuminating  Gas  anpl  Suction  Gas  Engines  103 

84.  Prices  of  Standard  Wrought  Iron  Pipe,  Casing, 

Wood  Pipe,  Riveted  Steel  Pipe.     Pipe  Fric- 
tion.   Cost  of  Tunnels,  Canals,  Excavations, 

Hydro-electric  Installations 107 

85.  Table  of  Prices  of  Various  Equipments 116 

CHAPTER  V.    VESTANCES 

86-87.     The  Time  Element  in  Vestance 126 

88-89.    Valuance 128 

90-97.    The  Steam  Engine 129 

98-99.     Change  Points 140 

loo-ioi .     Vestances  at  Full  and  Fractional  Loads  of  Steam 

Engines 141 


CONTENTS 

102.  Oil  Engines 

103.  Diesel  Engines I5° 

104-106.     Induction  Motors J52 

107.  Generators IS^ 

108.  Comparison  of  Power  Units i58 

109.  Vestances  at  Full  and  Fractional  Loads 159 

110-115.     Centrifugal  Pumps 162 

116-117.    Vestances  of  Pipe *74 

118-121.     Determination   of   Velocity   in   Pipe   for   Best 

Financial  Efficiency i?7 

122.  Oregon  Rates.    Vestances  of  Induction  Motors 

with  the  Oregon  Rates 184 . 

CHAPTER  VI.    UNIT  COST  DETERMINATION 

123.  Unit  Cost  of  Service 187 

124.  Time  Element  in  Service  Costs 187 

125-127.     Change  Points.    Unit  Cost  with  Constant  Load, 

and  Nearly  Constant  Load 188 

128-131.     The  Two-Part  Load 200 

132.     The  Three  Part  Load.    The  w-part  Load 210 

133-137.     Service  Modulus 213 

CHAPTER  VII.    DETERMINATION  OF  SIZE  OF  SYSTEM  FOR 
BEST  FINANCIAL  EFFICIENCY 

138.  Variable  Operating  Cost 222 

139.  Total  Production  Cost 224 

139.  Determination    of   Equation    of   Actual   Load 

Curve 224 

140.  Analysis  of  Type  of  System  for  Heat  Transmis- 

sion   232 

140.  Determination  of  Point  of  Best  Financial  Effi- 

ciency    232 

CHAPTER  VIII.    DETERMINATION  OF  TYPE  AND  SIZE  OF  UNITS 

141.  Stand-by  Units 242 

142.  Determination  of  Cost  of  Service  of  a  Unit  at 

Fractional  Loads 243 


x  CONTENTS 

143.  Determination  of  Cost  of  Service  of  a  Plant  at 

Fractional  Loads 244 

144-145.  Determination  of  the  Number  of  Units  in  a  Plant 

for  Best  Financial  Efficiency 246 

146-147.     Design  of  Plants  for  Best  Financial  Efficiency.     255 

INDEX 269 


FINANCIAL  ENGINEERING 


CHAPTER   I 
INTRODUCTION 

1.  In  designing  a  system  for  the  generation  and  distribution 
of  power,  or  in  laying  out  a -factory  for  the  manufacture  of  a 
certain  article,  or  in  the  rendition  of  any  other  service,  the 
duties  of  the  engineer  are  twofold:  In  the  first  place,  he  must 
so  design  the  system  that  it  will  operate  with  reasonable  con- 
tinuity.   In  the  second  place,  he  must  so  design  the  system  that 
it  will  operate,  not  only  with  reasonable  economy,  but  with  the 
best  possible  economy.    In  any  system  already  built  and  in 
operation,  the  engineer  should  be  able  to  determine  the  exact 
unit  cost  of  production  of  the  service  as  well  as  determine  what 
parts  of  the  system  are  operating  with  good,  and  which  with 
low  economy,  so  that  all  leaks  may  be  closed.    In  the  following 
pages  we  shall  treat  the  definite  and  exact  solution  of  such 
problems,  both  in  theory  and  in  practical  application. 

2.  The  entire  problem'  of  financial  engineering,   naturally 
divides  itself  into  two  parts,  namely: 

(a)  Cost  segregration  and 

(b)  Cost  analysis. 

Cost  segregation  deals  with  the  proper  allocation  of  the 
various  items  of  expense,  and  their  division  and  arrangement  as 
best  suited  for  cost  analysis.  This  is  essentially  the  book- 
keeping phase  of  the  problem,  and  will  therefore  be  treated  in 
the  following  pages  only  where  absolutely  necessary  and  no  more. 

Cost  analysis  deals  with  the  utilization  of  the  data,  obtained 
from  cost  segregation,  together  with  all  available  engineering 
data,  for  the  purpose  of  reaching  conclusions  as  to  the  economy 


INTRODUCTION 


of  a  system  a$  ^1dieie.ora|i  part,  whether  the  system  is  proposed 
or  already  exists. 

3.  In    cost  analysis,   we  start  from    the    basis  that  cost 1 
governs  price;  competition  and  utility  commissions,  only  the 
profit  therein  contained.     It  is  evident  that  if  a  servant  corpo- 
ration sells  for  less  than  cost,  it  must  eventually  go  into  bank- 
ruptcy.    If  it  sells  just  at  cost,  the  service  will  be  rendered 
without  reward.    The  servant  company  must  therefore  sell  at 
cost  plus  a  certain  margin  called  profit.    We  can  therefore  say 
that  cost  plus  profit  equals  price  or  rate. 

So  also,  an  equipment  must  earn  its  costs  and  profits,  when 
in  operation,  for  certainly  it  cannot  do  so  when  idle. 

4.  Profit.  —  This  word  is  used  in  two  senses,  namely  that  of 
gross  profit  and  that  of  net  profit.    Thus  if  an  article  cost  us 
$100  and  we  sell  it  for  $130,  then  the  gross  profit  is  $30.    If  the 
cost  of  making  the  sale  is  $20,  then  the  net  profit  is  $10.    It  is 
evident  that  gross  profit  is  not  profit  at  all  for  it  includes  essen- 
tial cost.    When  we  use  the  term  profit  in  the  following  pages 
we  shall  mean  only  net  or  real  profit. 

It  should  be  borne  in  mind  that  a  finished  article  almost 
invariably  carries  not  merely  just  one  profit  but  usually  a 
whole  series  of  profits.  Thus,  for  example,  a  certain  amount  of 
iron  is  mined  at  a  cost  of  $i  to  which  is  added,  let  us  say,  a 
profit  of  10%,  making  the  price  $1.10.  It  is  converted  into 
steel  at  a  cost  of,  say,  $0.50,  making  the  cost  to  the  steel  pro- 
ducer $1.60.  To  this  a  profit  of  10%  is  added,  making  the  price 
of  the  steel  $1.76.  The  manufacturer  converts  this  steel  into  a 
machine  part  at  a  cost  of  $1.24,  making  a  cost  to  him  of  $3. 
To  this  he  adds  10%  profit,  making  the  price  $3.30.  Finally 
the  sales  organization  disposes  of  this  part  at  a  cost  of  $0.70, 
making  their  cost  $4.  To  this  they  likewise  add  10%  profit, 
making  the  final  selling  price  $4.40. 

Under  such  conditions,  we  would  have  a  total  cost  of 

i.oo  -f  0.50  +  1-24  -f  0.70  =  $3.44 
and  total  profits  of 

o.io  +  0.16  -f-  0.30  +  0.40  =  $0.96. 

1  Goldman,  Trans.  A.  I.  E.  E. 


BASIS  OF  RATES  3 

The  over-all  per  cent  of  profit  is  then 

,0.96  -r-  3.44  =  35.8%. 

5.  Value.  —  While  we  can  and  shall  determine  comparative 
values,  we  have  come  to  realize  that  "  value  "  has  no  meaning 
at  all  in  any  exact  sense,  for  the  reason  that  we  possess  no 
absolute  standard  of  measurement  by  which  to  measure  it. 
Furthermore  value  is  intensely  variable  with  circumstances. 
What,  for  example,  is  the  value  of  a  meal  to  one  that  has  just 
been  fed  as  compared  with  the  value  of  the  same  meal  to  a 
man  who  is  starving?    It  was  for  some  time  attempted  to  use 
the  value  of  service  as  the  basis  of  rates,  but  it  was  found  to  be 
synonymous  with  "  what  the  traffic  will  bear,"  a  method  of  • 
coining  money  from  the  wants  or  distress  of  others. 

6.  Basis  of  Rates.  —  Each  servant  corporation  must  recover 
from  its  employers,  i.e.,  customers,  all  its  costs,  to  which  is 
added   a   certain  per   cent  of  profit.     According  to  present 
arrangements,  it  is  not  reasonable  to  expect  a  service  as  a 
whole,  to  be  rendered  without  profit.    And  what  applies  to  the 
service  as  a  whole,  applies  equally  well  to  each  and  every  single 
item  of  the  service.    It  is  as  unreasonable  to  expect  any  item 
of  the  service  to  be  rendered  without  profit  as  to  expect  the 
service  as  a  whole  to  be  so  performed.     But  while  it  is  com- 
paratively easy  to  determine  if  the  service  as  a  whole  is  profit- 
able, by  little  more  than  consulting  bank  balances,  it  is  entirely 
another  matter  to  do  this  for  any  single  item  of  service. 

While  it  is  unreasonable  to  expect  a  servant  company  to 
render  a  service  or  any  item  thereof  without  profit,  it  is  equally 
unreasonable  to  expect  one  customer  or  any  one  group  of  cus- 
tomers, to  pay,  besides  the  cost  burden  that  they  place  on 
the  undertaking,  more  than  their  pro-rata  of  the  profit.  Nor 
would  it  be  equitable  to  make  one  group  of  customers  pay, 
besides  their  cost  burden  and  pro-rata  of  profit,  the  cost  burden 
of  any  other  group  of  customers.  In  other  words,  it  is  not  just 
to  take  the  cost  or  profit  burden,  or  any  part  thereof,  off  of 
one  group  of  customers  and  place  it  on  another  group.  Although 
in  this  way,  the  profit  of  the  undertaking  as  a  whole  could  be 
maintained  unaltered,  it  would,  nevertheless,  be  nothing  short 


4  INTRODUCTION 

of  discrimination  in  favor  of  one  group  as  against  another. 
For  this  reason,  it  has  been  held  illegal  according  to  decisions 
of  the  U.  S.  Supreme  Court. 

It  is  not  to  be  inferred  from  the  above  that  each  servant 
company  should  receive  the  same  per  cent  of  profit.  For  under 
such  conditions,  there  would  be  no  reward  for  efficiency,  no 
incentive  to  engage  the  best  engineering  and  administrative 
talent,  and  no  punishment  in  the  way  of  reduced  incomes  for 
poor  economy. 

7.  Since  cost  governs  price,  the  entire  problem  of  rate  or 
price  determination  is  primarily  that  of  engineering;    that  of 
design  for  maximum   economy  is  entirely  so.     Knowing  the 
exact  unit  cost,  the  price  made  therefrom  is  a  matter  of  judg- 
ment, for  it  is  founded  on  fact.    But  a  price  or  rate  made  with- 
out knowledge  of  the  unit  cost,  is  an  unadulterated  matter  of 
guesswork,  if  not  worse. 

Very  often  one  company  will  set  its  prices  in  accordance  with 
those  of  its  competitors.  That  is  one  company  will  copy  the 
prices  of  its  competitors  and  use  these  prices  as  its  own.  This 
looks  right  on  the  ground  that  "  competition  regulates  the 
price/'  But  it  is  not  right  because  competition  does  not 
regulate  the  price  but  only  at  best  what  we  may  demand  in 
profit.  If  the  competitor's  prices  are  less  than  our  costs,  then 
we  must  obtain  a  higher  price,  turn  our  attention  to  some  other 
line  of  endeavor  or  else  go  broke. 

8.  In  the  problem  of  rate  determination  for  public  utilities, 
altogether  too  much  stress  has  been  laid  on  the  various  deci- 
sions of  our  state  courts,  in  spite  of  the  fact  that  these  deci- 
sions have  been  exceedingly  contradictory.     Nor  could  these 
decisions  help  being  so,  for  the  problems  to  be  solved  are  those 
of  engineering  and  not  law.    The  latter  is  therefore  helpless  in 
reaching  a  solution  until  engineering  shows  the  way.    In  other 
words  engineering  must  lead  in  these  matters  and  determine 
the  correct  solutions  to  which  the  law  must  conform. 

Courts  base  their  decisions  on  the  expert  opinion  of  engineers 
in  so  far  as  these  opinions  are  based  on  exact  scientific  knowl- 
edge. Beyond  that  the  courts  have  as  much  right  to  guess  as 


DIVISION  OF   SERVICE  5 

anyone.  If  the  opinion  of  the  experts  were  crude  and  far  from 
the  truth,  then  the  decisions  based  thereon  are  bound  to  be 
equally  bad. 

It  must  be  borne  in  mind  that  bad  decisions  of  our  courts, 
going  as  they  do,  counter  current  to  the  trend  of  natural 
development  of  human  society,  are  a  great  hindrance  to  progress, 
while  good  ones  are  a  great  aid.  While  bad  decisions  cannot 
prevent,  they  most  assuredly  do  delay  advancement.  After  all 
it  must  be  borne  in  mind  that  knowledge  is  all  one  and  indi- 
visible. We  have  hypothetical  divisions,  not  real  ones.  One 
so-called  branch  of  knowledge  can  no  more  live  alone  and 
separate  from  all  other  knowledge  than  the  hand  or  the  heart 
can  exist  separate  from  the  whole  living  being.  A  discovery  or 
advancement  in  engineering  or  a  good  decision  in  law  both 
aid  one  another  in  that  both  aid  in  the  progress  of  human 
society 

9.  As  we  have  pointed  out  above,  there  are  primarily  two 
divisions  to  our  entire  problem,  namely  that  of  cost  segregation 
and  cost  analysis,  as  applied  to  (a)  cost  determination  and  (b) 
design  for  best  economy.  Each  charge  growing  out  of  the 
rendition  of  a  service  must  be  recorded,  charged  against  the 
proper  account,  while  mixed  charges,  applying  to  more  than 
one  item,  must  be  properly  apportioned.  The  apportioning 
of  mixed  items  of  expense  is  a  task  requiring  a  broad  knowledge 
of  engineering  and  perfect  common  sense.  Cost  segregation 
must  be  made  in  accordance  with  the  demands  of  cost  analysis. 
For  the  cost  segregation  is  of  no  value  in  and  of  itself,  and  is 
useful  only  as  an  absolutely  necessary  basis  for  cost  analysis. 
Cost  segregation  alone,  where  no  further  use  is  made  of  it, 
as  is  so  often  done,  represents  just  so  much  useless  and 
undigested  data. 

Drawing  conclusions  from  data  is  the  part  of  cost  analysis. 
It  is  the  business  of  determining  exact  unit  cost  for  actual  load 
conditions  under  which  the  'system  operates,  as  well  as  the 
aggregate  effect  of  the  load  characteristics  of  each  unit  of  the 
system.  In  the  problem  of  design,  the  field  is  still  broader. 
The  system  is  not  yet  fixed,  but  is  to  be  determined  so  that, 


6  INTRODUCTION  . 

under  actual  or  anticipated  load  conditions,  the  greatest  econ- 
omy will  be  attained.  This  part  of  the  problem  may  be 
subdivided  into: 

(a)  The  determination  of  the  type  and  size  of  a  system  for 
best  economy 

(b)  The  determination  of  the  type  and  size  of  the  units  of  a 
system  for  best  economy 

(c)  The  determination  of  unit  and  total  annual  production 
cost. 

10.  The  service  that  is  rendered  by  every  servant,  whether 
of  a  private  or  public  character,   may  be  divided  into  four 
primary  parts,  namely: 

(a)  Collection, 

(b)  Production, 

(c)  Transmission  (Transportation)  and 

(d)  Distribution. 

A  railway  service,  for  example,  consists  in  the  collection  of 
freight,  its  transportation  and  distribution,  production  being 
zero  as  nothing  is  produced  as  such.  A  power  service  consists 
in  the  collection  (storage)  of  water  for  power,  the  production 
of  power,  its  transmission  and  distribution.  Again  an  engine 
factory  service  consists  in  the  collection  of  materials  of  manu- 
facture, the  production  of  the  engines,  their  transportation  and 
distribution. 

The  actual  rendition  of  one  of  these  divisions  of  a  complete 
service  may  be  delegated  to  an  agent.  Thus,  for  example,  in 
the  case  of  an  engine  factory,  the  actual  transportation  of  the 
engines  is  almost  invariably  left  to  a  railway  company,  who,  by 
specializing  for  this  particular  division  of  service,  can  render 
it  at  an  incomparably  lower  cost.  So  also  the  actual  distribu- 
tion of  the  engines  may  be  left  to  a  sales  agency.  Such  is  in 
fact  very  commonly  done. 

11.  All  cost  pertaining  to  the  rendition  of  a  service  may  be 
divided  into  two  parts,  namely: 

(a)  Fixed  charges,  and 

(b)  Operating  costs. 


FIXED   AND   OPERATING  COSTS  7 

Fixed  charges  are  all  those  costs  which  continue  when  the 
operation  of  a  system  is  discontinued.  Thus,  interest,  taxes, 
and  the  like  continue  whether  the  plant  is  idle  or  in  use,  i.e., 
productive  or  nonproductive.  Thus  the  total  annual  costs  of 
reserve  units,  while  standing  idle,  is  merely  the  total  annual 
fixed  charges  of  these  units,  the  operating  costs  being  zero. 
Such  stand-by  costs  must  be  considered  an  insurance  against 
discontinuity  of  service  and  needs  be  as  carefully  determined  for 
best  economy  as  any  other  factor  of  the  system. 

The  operating  cost  consists  of  all  costs  over  and  above  those 
of  fixed  charges.  The  costs  of  fuel,  oil,  water  (materials  con- 
sumed) ,  attendance,  and  the  like  are  clearly  all  operating  costs. 
But  the  salary  of  the  engineer  may  be  partly  an  operating  and 
partly  a  capital  cost.  When  the  time  and  skill  of  the  engineer 
is  spent  on  the  running  of  the  system,  or  its  maintenance,  or 
repair,  it  is  an  operating  cost.  But  when  it  is  spent  on  the 
design  of  extensions  or  their  construction,  the  salary  must  be 
charged  to  capital  and  bear  fixed  charges  as  any  other  item  of 
investment.  The  same  holds  for  supervision  and  the  like  of 
what  may  be  called  the  overhead  services. 

We  may  therefore  segregate  our  costs  as  follows: 

(1)  Fixed  charges: 

(a)  Interest  and  rents, 

(b)  Depreciation, 

(c)  Taxes, 

(d)  Insurance. 

(2)  Operating  costs: 

(a)  Materials  consumed,  power,  etc., 

(b)  Attendance, 

(c)  Maintenance, 

(d)  Repair. 

12.  Ordinarily,  undertakings  are  divided  into  private  and 
public  utilities.  But  the  production  of  steel,  clothes,  food  or 
the  like  is  as  much  a  public  necessity  as  transportation  or  power 
service.  The  above  classification,  though  legal,  is  therefore 


8  INTRODUCTION 

both  arbitrary  and  temporary.  Instead  of  the  above,  we  shall 
classify  all  undertakings  into  individual  and  integral  under- 
takings, as  well  as  mixed. 

An  integral  system  is  one  that  is  designed  to  serve  one  and 
only  one  —  one  community  or  one  given  district  and  all  of 
that  community  or  district.  Such  a  system  is  designed  for  the 
general  needs  of  the  community  that  it  serves  and  not  the 
particular  needs  of  certain  individuals  of  that  district,  except 
in  so  far  as  service  connections  are  needed.  These  needs  can 
be  closely  determined  from  previous  experience  with  similar 
communities. 

An  individual  system  is  one  that  is  designed  to  meet  part  or 
all  of  the  needs  of  certain,  definite  individuals.  The  distinc- 
tion may  be  made  clear  by  an  illustration.  Thus  a  telephone 
or  power  system  is  distinctly  an  integral  system.  You,  indi- 
vidually, do  not  pay  any  charge  unless  you  are  actually  receiv- 
ing service  from  the  undertaking,  although  the  necessary  costs 
and  profits  for  operating  the  undertaking  are  collected  from 
the  community,  in  particular  from  that  part  of  the  community 
which  does  receive  service.  The  undertaking  is  'ready'  to  serve 
you,  but  that  costs  you  nothing.  All  public  utilities  and  most 
private  companies  are  integral  undertakings. 

On  the  other  hand  if,  let  us  say,  six  men  join  to  put  in  a 
common  pumping  system,  we  have  an  individual  system,  for 
the  size  and  type  of  the  system  are  chosen  to  suit  the  needs  of 
just  these  particular  six  men  and  no  more  or  less.  In  such  a 
system  the  readiness  to  serve  must  be  paid  for  whether  service 
is  actually  taken  or  not. 

Example  i :  Four  men,  each  owning  40  acres  of  land,  put  in  a 
common  pumping  system  costing  $4000.  One  man  uses  an 
average  of  5  acre  feet  of  water  per  acre  per  year,  the  second  4, 
the  third  3,  and  the  fourth  no  water  at  all.  If  the  fixed  charges 
on  the  system  are  12%,  and  the  operating  costs  are  50  cents 
per  acre  foot,  determine  the  annual  charges  that  must  be 
assessed  against  each  man. 

Solution:  The  fixed  charges  are 

4000   X    12%    =   $480. 


INDIVIDUAL  AND  INTEGRAL   SYSTEMS  9 

Since  the  system  is  designed  to  serve  each  man  equally  and 
as  each  man  owns  the  same  amount  of  land,  the  fixed  charges 
will  be  divided  equally,  each  man  paying  one-fourth,  or  $120, 
per  year. 

The  total  water  pumped  is 

5  X  40  =  200  acre  feet. 
4  X  40  =  1 60    "      " 
3  X  40  =  120    "      " 
o  X  40  =  OOP     "      " 

Total  480  of  water, 

costing 

480  X  .50  =  $240. 

This  amount  may  be  considered  as  divided  into 
5+4  +  3+0  =  12  parts. 

Of  this  the  first  man,  using  5  acre  feet,  pays  in  operating 
costs 

T\  X  $240  =  $100; 
the  second  pays 

T42  X  $240  =  $80; 
the  third  pays 

i32-  X  $240  =  $60; 

and  the  fourth,  using  no  water,  pays  no  operating  cost. 

Evidently,  then,  the  first  man  must  pay  a  total  in  fixed  and 
operating  charges  of       120  +  100  =  $220; 
the  second,  120  +    80  =  $200; 

the  third,  120  +    60  =  $180; 

and  the  fourth 

1 20.00  +  oo.oo  =  $120. 

Although  the  fourth  man  receives  no  service,  he  must  pay 
his  pro-rata  of  fixed  charges,  his  "  readiness  to  serve  "  charge. 
For  were  he  not  in  the  system,  the  size  and  thus  the  total  cost 
and  fixed  charges  of  the  system  could  be  proportionately  re- 
duced. This,  however,  holds  true  only  for  individual  systems. 

13.  The  government  has  always  exercised  more  or  less  con- 
trol over  what  are  now  classed  as  public  utilities.  But  not  until 
special  commissions  were  formed  did  this  control  begin  to  be 


io  INTRODUCTION 

effective.  Even  so,  the  commissions  have  been  weak  and 
unreasonable  in  many  respects.  But  this  must  be  expected 
to  a  certain  extent  in  the  incipient  exercise  of  this  authority. 
And  it  will  continue  until  the  commissions  are  composed  of 
experts  and  experts  only. 

It  is  now  clearly  established  that  actual,  real  costs  must  be 
used  as  the  basis  of  rates  in  such  undertakings,  and  not  the 
hypothetical  "  value  of  service,"  or  the  piratical  "  what  the 
service  will  bear  "  basis.  "For,"  says  Justice  Hughes  in  the  de- 
cision of  the  United  States  Supreme  Court  in  the  North  Dakota 
coal  case,  "  where  it  is  established  that  a  commodity  .  .  .  has 
a  rate  imposed,  which  would  compel  the  carrier  to  transport 
it  ...  virtually  at  cost,  and  thus  the  carrier  would  be  denied 
a  reasonable  reward  for  its  (that  particular)  service  ...  it 
must  be  concluded  that  the  state  (commission)  has  exceeded 
its  authority."  This  decision,  at  a  single  sweep,  landed  most  of 
the  work  of  our  inexpert  state  commissions  in  the  waste  basket. 

14.  Competition  is  industrial  disorganization.  Under  such 
conditions  no  service  can  be  rendered  economically,  for  it 
involves  the  wasteful  expenditure  of  capital  and  labor.  The 
formation  of  the  so-called  trusts  is  a  natural  attempt  to  evade 
such  conditions.  Under  competitive  conditions,  the  public 
gets  poor  and  expensive  service  while  the  returns  on  the  under- 
taking are  often  subnormal.  It  is  true  that  in  some  cases  better 
and  cheaper  service  has  been  rendered  under  competitive  con- 
ditions. But  this  was  due  either  to  lack  of  ability  of  those  in 
control  of  the  first  undertaking,  or  abnormally  large  profits 
collected,  due  either  to  lack  of  authority,  or  ability,  of  the 
commission  or  both.  Commissions  should  have  the  ability 
to  know  what  is  right,  the  courage  to  do  what  is  right,  and  the 
authority  to  execute  it.  We  emphasize  this  because  it  seems 
probable  that  the  authority  of  the  commissions  will  be  extended 
eventually  to  all  integral  systems.  Competition  is  always  the 
short  cut  to  higher  costs. 

Economists  have  stated  that  there  was  such  a  thing  as 
the  law  of  demand  and  supply,  which  automatically  regu- 
lates prices.  Even  the  most  elementary  study  shows  that 


COMPARATIVE  VALUES  n 

there  is  no  such  law.  In  the  first  place  the  demand  for  any 
necessity  is  practically  a  constant  but  the  amount  used  will 
depend,  not  so  much  upon  the  price,  as  upon  the  wealth  (pur- 
chasing ability)  of  the  consumer.  In  the  second  place,  while 
it  is  true  that  the  prices  vary  with  the  supply,  this  is  due  to 
the  unregulated,  unscientific  methods  of  production,  giving 
us  alternate  waves  of  oversupply  and  undersupply,  alternate 
waves  of  waste  and  want.  If  there  is  such  a  law,  it  is  the  law 
of  the  wilds,  where  they  stuff  in  summer  and  starve  in  winter. 
Certainly  it  is  not  the  law  of  intelligent  production. 

Cooperation  and  not  competition,  work  and  not  war  and 
waste  is  the  keystone  of  our  modern  social  structure.  This 
accounts  for  the  condensation  of  small  industrial  units  into 
fewer  large  one,  for  the  formation  of  the  labor  unions,  fraterni- 
ties and  the  like.  Much  as  capitalism  has  been  condemned, 
one  thing  certainly  may  be  said  for  it;  that  it  compelled 
cooperation  and  coordination  amongst  workers  long  before 
there  would  have  been  voluntary  cooperation  and  in  so  doing 
has  served  to  advance  civilization.  But  with  the  coming  of 
voluntary  association  and  cooperation,  compulsory  association 
and  compulsory  cooperation  will  be  little  tolerated. 

15.  The  comparative  value  of  an  undertaking  is  in  direct 
proportion  to  its  size  and  economy  of  rendering  the  service. 
There  are  even  yet,  however,  a  number  of  different  "  methods  " 
for  determining  the  so-called  " value"  of  an  undertaking. 
These  are  based  on  (i)  Original  Cost,  (2)  Replacement  Cost, 
(3)  Market  Value  of  Outstanding  Liabilities,  and  finally  that 
based  on  (4)  the  Capitalization  of  Earnings. 

The  Original  Cost  includes  not  only  the  money  spent  on  the 
undertaking  at  its  inception,  to  bring  it  into  being,  but  also  all 
money  spent  for  extensions  and  improvements,  except  in  so  far 
as  such  may  be  chargeable  to  depreciation.  Original  cost  will 
however,  not  give  the  comparative  value  of  an  undertaking  for, 
according  to  this  basis,  the  more  a  given  undertaking  cost,  the 
greater  would  be  its  worth.  It  is  evident  then  that  original 
cost  will  lead  to  the  comparative  value  only  if  we  take  into 
consideration  most  fully  the  economy  of  the  undertaking.  In 


12  INTRODUCTION 

other  words  we  must  determine  exactly  how  well  it  serves  the 
purpose  for  which  it  was  constructed. 

The  Replacement  Cost  is  the  amount  of  money  that  would 
have  to  be  expended  now,  or  at  some  other  time  subsequent 
to  the  original  formation  of  the  undertaking,  to  bring  it  into 
being.  Replacement  cost  includes,  besides  the  original  cost, 
the  depreciation  of  this  sum  of  money  in  purchasing  power 
during  the  interim  and  therefore  tends  to  give  the  undertaking 
a  speculative  value. 

The  "  market  value"  of  the  outstanding  liabilities  of  a  com- 
pany should  give  some  information  as  to  the  comparative  value 
of  an  undertaking,  but,  with  rare  exception,  they  most  cer- 
tainly do  not.  The  real  value  is  usually  quite  dilute.  There 
are  many  reasons  for  this,  but  fundamentally,  whenever  a 
corporation  has  the  power  to  create  an  artificial  market,  then 
these  " market  values"  represent  nothing  but  the  ability  of 
the  companies  along  these  lines  of  manipulation. 

Attempting  to  find  the  " value"  of  an  undertaking  from  its 
earnings,  so  as  to  determine  the  earnings  from  the  " value"  so 
found,  is  like  trying  to  find  the  end  of  a  circle.  A  company 
might,  for  example,  be  guilty  of  exploiting  the  community 
that  it  serves.  If  the  " value"  of  the  undertaking  under  such 
conditions  is  determined  from  its  earnings,  it  would  give  such  a 
concern  a  "vested  right"  in  such  exploitation. 

It  must  be  clearly  kept  in  mind  in  discussing  the  above  that 
the  word  "  value  "  has  no  definite  meaning.  In  view  of  this,  the 
purposelessness  of  the  above  "methods"  becomes  more  evident. 

In  order  to  illustrate  the  enormous  discrepancy  in  deter- 
mining so-called  "value"  by  the  various  methods  above,  we 
give  below  some  of  these  "values"  for  one  certain  company. 

(a)  Market  "Value"  as  per  outstanding  liabilities,  $72,000,000 

(b)  Present  "Value"  as  per  utility  commission.  .  .  36,000,000 

(c)  Present  "Value"  as  per  assessor 15,000,000 

(d)  "Value"  as  per  capitalized  earning 20,000,000 

(e)  Estimated  real  present  value 16,000,000 

The  above  speaks  for  itself.    A  guess  is  scientific  in  comparison. 


PROFITS  13 

16.  It  took  the  great  war  to  bring  to  full  realization  the  fact 
that  gold  is  not  a  necessity,  that  necessities  are  not  measured 
by  gold,  but  gold  by  necessities.    Under  the  present  era  (1918) 
of  "high  prices,"  it  takes  no  more  sacks  of  potatoes  or  suits  of 
clothes  to  build  a  house  than  under  normal  times  although  it 
takes  twice  as  much  gold.     It  is  not  the  price  of  all  commodi- 
ties that  has  increased,  but  the  price  of  gold  that  has  decreased. 
It  is  far  more  sensible  to  acknowledge  this  view  than  to  take 
the  opposite  one  that  everything  has  changed  but  gold.    When, 
for  example,  the  price  of  wheat  was  set  at  $2  per  bushel,  it  was  in 
reality  not  the  price  of  wheat  that  was  set  at  all,  but  the  price 
of  gold  that  was  set.    This  allowed  a  50%  depreciation  in  the 
price  of  gold  and  had  this  not  been  done,  its  depreciation  would 
have  been  far  greater,  far,  probably,  below  its  cost  of  production. 

That  is  why  the  payment  of  wages  in  gold  has  proved  unsatis- 
factory, because  the  " doubling"  of  wages  under  such  conditions 
meant  no  raise  at  all.  But  were  they  based  on  necessary  com- 
modities, wages  would  at  once  be  stabilized. 

17.  Profits   have   been   a   fruitful   source   of   political   and 
economic  discussion.    The  view  generally  taken  is  that  profit 
is  the  reward  that  the  servant  gets  fpr  rendering  a  service, 
notwithstanding  that  the  wages,  that  is  the  total  costs,  do 
pay  for  this  service.    The  peculiar  anomaly,  therefore,  exists 
that,  if  this  servant  is  a  human  being,  he  gets  only  wages  for 
rendering  the  service,  but,  if  the  servant  is  an  undertaking,  it 
gets  not  only  full  wages,  but  profit  besides.    Looking  at  it 
from  the  standpoint  of  the  purchaser,  i.e.,  the  employer  of  the 
service,  what  does  he,  the  purchaser  (employer) ,  get  for  the  profit 
that  he  may  pay  to  the  servant?    It  is  evident   that,  if  the 
costs  (or  wages)  are  the  full  reward  for  the  service  rendered,  the 
employer  (customer)  obtains  absolutely  nothing  for  this  profit 
that  he  pays  above  the  cost  (total  wages)  of  a  commodity. 

18.  It  is  evident  from  the  above  that  all  business  organiza- 
tions, whether  we  choose  to  speak  of  them  as  companies,  under- 
takings, utilities,  corporations,  or  the  like,  have  but  one  purpose 
and  that  is  the  rendition  of  a  service  for  others.     They  are 
evidently  just  servant  organizations  and  we  shall  in  these  pages 


14  INTRODUCTION 

speak  of  them  on  occasion  as  servants.  The  relation  of  a 
customer,  patron,  or  purchaser  to  an  organization  or  individual 
from  whom  he  obtains  a  service  (or  commodity)  is  that  of 
employer  to  employee  in  that  he  employs  the  organization 
to  render  the  service.  The  service  of  such  an  organization  is 
rendered  by  the  organized  efforts  of  individual  servants.  The 
relation  of  the  organization  to  the  individual  servants  is  that 
of  employer  to  employee.  But  the  relation  of  the  public  to 
a  public  utility  is  (or  should  be)  that  of  Master  to  Servant. 

In  the  case  of  an  individual  servant,  there  is  no  question 
but  that  the  price  equals  the  total  wages,  or  cost  of  the  service. 
But  in  the  case  of  an  organization  of  Servants,  as  a  corporation, 
the  price  is  greater,  often  much  greater,  than  the  wages  (costs) , 
and  this  excess  or  profit  does  not  go  to  the  individual  servants  of 
that  organization  but  to  others,  who  render  no  service  whatever. 

It  must  be  borne  in  mind  clearly  that  all  costs  of  a  given 
service  are  exactly  equal  to  all  wages  associated  with  that 
service.  Usually  such  costs  are  spoken  of  as  wages  proper  and 
cost  of  materials.  But  the  cost  of  the  materials  themselves  is 
only  wages,  so  that,  in  fact,  cost  equals  total  wages.  So-called 
overhead  expenses  may  be  divided  into  salaries  paid  for  essen- 
tial service  rendered  and  thus  come  under  wages,  rents  which 
in  the  main  come  under  profits  and  so  forth.  Wages  as  used 
herein  is  understood  to  mean  all  moneys  paid  for  essential 
service  rendered.  Iron  ore,  for  example,  has  no  cost,  being  a 
natural  resource.  But  a  finished  machine  has  for  its  cost  —  as 
distinguished  from  its  price  —  all  moneys  paid  for  essential 
service  of  whatever  kind,  in  the  mining  of  the  ore,  its  smelting, 
transportation,  etc.,  and  its  manufacture  finally  into  a  finished 
machine.  The  cost  of  such  a  machine,  or  any  other  material 
we  may  buy,  is  the  sum  total  of  all  wages  associated  with  its 
production  and  manufacture. 

The  best  member  of  this,  or  any  other  country,  is  the  one 
that  renders  the  community  the  best  and  most  service.  It 
certainly  is  not  the  " successful  man"  who,  instead  of  serving 
the  community,  has  merely  proven  his  ability  to  take  and 
gather  the  most  unto  himself,  instead  of  giving  the  most. 


CHAPTER  II 
COST  SEGREGATION 

Interest.  Depreciation  and  Appreciation.  Taxes.  Insurance.  Materials 
Consumed.  Attendance.  Maintenance  Repair.  Life  of  Structures 
and  Equipment.  Stock.  Storage.  Sale  and  Distribution. 

19.  As  pointed  out  in  the  first  chapter,  all  production  costs 
may  be  divided  into  fixed  charges  and  operating  costs.    The 
fixed  charges  are  divided  into  the  following  divisions: 

(a)  Interest  and  rents, 

(b)  Depreciation, 

(c)  Taxes, 

(d)  Insurance, 

while  the  operating  costs  are  divided  into : 

(a)  Materials  consumed,  power,  etc., 

(b)  Attendance, 

(c)  Maintenance, 

(d)  Repair. 

20.  Interest.  —  The  price  of  any  service,  or  commodity,  is 
made  up  of  just  two  items,  namely:   the  cost,  or  total  wages 
and  salaries,  and  the  profit.     To  which  of  these  does  interest 
belong?    It  is  a  nonproductive  charge,  not  being  paid  to  the 
individual  servants  of  the  organization  but  to  some  one  else  who 
renders  no  service.     It  is  evidently  a  profit  charge  and  is  so 
treated  by   both   federal   and   state  commissions.      So-called 
public   utilities  are  allowed   all   their  profits  in  the  form  of 
interest  on  the  " capital"  invested,  and  no  profit  is  allowed 
in  any  other  form.    This  is  from  the  actual  standpoint  of  the 
customer  (employer)  of   the  service.     An  organization  using 

15 


16  COST  SEGREGATION" 

borrowea  money  sees  this  interest  as  a  cost,  and  we  so  treat 
it  in  the  following  pages. 

Interest  is  charged  on  all  money  invested.  This  consists  of 
the  price  paid  for  all  machinery,  materials,  and  wages,  as  well 
as  unavoidable  losses,  such  as  loss  of  time  during  construction 
due  to  inclement  weather,  loss  of  tools,  cost  of  temporary 
structures  and  the  like. 

The  rate  of  interest  varies  greatly,  usually  from  two  to 
eight  per  cent.  Governments  normally  pay  two  to  two  and  a 
half  per  cent,  large  industrial  undertakings  four  to  six  per  cent, 
while  small  firms  and  individuals  of  very  moderate  wealth  pay 
seven  to  eight  per  cent.  It  is  not  clear  what  determines  the 
rate  of  interest  unless  it  is  "all  that  the  traffic  will  bear."  It  is 
often  said  that  the  "risk"  determines  the  rate  of  interest,  so 
that  the  greater  the  "risk"  the  greater  the  rate  of  interest. 
In  other  words,  the  smaller  and  weaker  the  firm  (or  individual) 
the  greater  the  profit  burden  it  must  bear.  But  as  #  matter 
of  fact,  the  greater  the  rate  of  interest,  the  greater  the  "risk" 
naturally  becomes.  So  this  is  evidently  not  the  basis  of  interest 
but  rather  the  want  or  distress  of  the  user.  As  a  matter  of 
fact,  money  cannot  produce  money  any  more  than  a  reservoir 
can  produce  water. 

In  general,  what  applies  to  interest,  which  is  merely  the 
.rent  paid  for  capital,  applies  equally  well  to  rent  paid  for  the 
loan  of  any  other  commodity,  in  so  far  as  the  rent  charged 
for  any  such  commodity  exceeds  the  actual  costs  associated 
with  the  making  and  collecting  of  the  loan,  the  maintenance 
of  the  commodity  in  its  original  condition  and  the  depreciation 
associated  with  this  article. 

Net  profit,  net  interest  and  net  rent  are  all  alike  in  that 
they  are  all  net  profits,  i.e.,  charges  for  which  no  services  are 
rendered.  The  fact  that  the  profits  are  paid  to  others,  besides 
the  owners  of  the  organization,  does  not  reduce  the  burden 
that  the  customer  must  bear. 

21.  Depreciation  and  Appreciation.  —  As  things  become 
older,  their  comparative  value  changes.  If  the  comparative 
value  decreases,  we  have  depreciation,  while  if  it  increases, 


DEPRECIATION  17 

we  have  appreciation,  or  negative  depreciation.  A  given  item 
depreciates  due  to  its  tendency  to  become 

(a)  obsolete, 

(b)  inadequate, 

(c)  useless. 

What  we  speak  of  as  land  has  really  two  meanings,  namely, 
the  land  itself  and  location.  For  agricultural  purposes  we  buy 
the  land  itself,  and,  unless  we  fertilize  it,  it  depreciates  rapidly. 
For  marketing  (selling)  purposes  we  buy  location  and  this 
invariably  appreciates  with  increase  in  population.  The  price 
or  rent  of  locations  varies  with  the  serviceability  of  the  loca- 
tions to  the  public  (the  employer).  By  thus  varying  the 
price,  the  location  is  made  to  serve  the  owner  (servant)  instead 
of  the  public  (the  employer).  Locations  thus  become  the  toll- 
gates  of  industry,  though  not  the  only  tollgates. 

22.  Obsolescence.   Progress  in  engineering  results  in  im- 
provements that  yield  either  increased  efficiency  or  reduced 
first  cost.    For  this  reason,  a  unit  that  we  have  in  service  loses 
its  comparative  value  or  becomes  obsolete.    When  the  efficiency 
of  new  apparatus  has  been  so  far  improved  that  it  would  pay 
to  discard  the  unit  in  use  entirely,  then  the  unit  in  use  has 
become  completely  obsolete   and   its   comparative  value  zero 
or  even  much  less,  even  though  it  is  otherwise  in  perfect  run- 
ning condition.    Thus,  for  example,  a  uniflow  engine  will  give 
as  good  efficiency  as  a  triple  expansion  engine  while  costing 
only  40%  as  much.    Had  one  just  purchased  a  triple  expansion 
engine,  when  the  uniflow  appeared,  60%  of  the  purchase  price 
of  the  former  would  have  gone  into  depreciation  at  once,  for 
the  reason  that  the  same  service  could  be  performed  equally 
well  with  this  reduction  in  capital  outlay. 

23.  Inadequacy.  —  It   often   happens   in   a   new   and   fast 
growing  business  that  certain  items  of  the  system  may  be 
replaced  to  advantage  though  not  obsolete.    Thus  in  a  power 
plant,  it  often  is  an  advantage  to  replace  a  number  of  smaller 
units,  however  modern,  that  have  accumulated  during  growth, 
with  one  large  unit.    Again  in  the  case  of  a  railway  system, 


i8  COST  SEGREGATION 

the  traffic  may  increase  so  that  a  single  track  bridge  is  incapable 
of  handling  the  traffic,  although  it  is  still  in  good  condition. 
In  such  cases  the  item  affected  has  become  inadequate.  It  may 
be  argued  that  a  larger  unit  should  have  been  installed  in  the 
first  place  and  in  some  cases  that  would  have  been  right,  but 
more  often  this  would  have  resulted  in  reduced  economy  due 
to  too  early  an  outlay  of  capital. 

Only  when  the  plan  of  installing  smaller  units  first  and 
replacing  them  with  larger  units  later  results  is  better  economy 
than  having  installed  the  larger  units  at  the  start,  have  we  a 
case  of  inadequacy  of  equipment.  When  conditions  are  reversed 
we  have  a  case  of  inadequacy  of  the  management. 

24.  Uselessness.  —  A  system  or  part  thereof  becomes  use- 
less (without  use)  when  the  service  for  which  it  was  intended  is 
no  longer  demanded.    Thus  a  temporary  structure,  used  during 
construction,  becomes  useless  and  in  fact  often  a  nuisance,  when 
the  construction  is  finished.    Again  a  logging  railroad  may  be 
built  into  a  certain  district.    When  the  logging  is  finished,  the 
railroad  becomes  useless,  though  some  parts  of   it  may  be 
salvaged. 

25.  Fundamentally,   the   object   of   cost   calculation  is   to 
maintain  the  capital  intact,  neither  permitting  it  to  increase 
nor  decrease.    For  this  reason,  it  is  necessary  to  set  aside  each 
year  a  certain  sum  of  money,  equal  to  the  amount  that  the 
system  has  depreciated  in  that  time.    This  money,  so  set  aside, 
forms  the  Depreciation  Reserve  or  Sinking  Fund,  the  former 
term  being  preferable.    This  depreciation  is  an  essential  part 
of  the  cost  of  the  production  of  the  service  being  rendered. 
The  amount  of  the  depreciation  reserve  laid  by  should  be  such 
that,  at  any  time  that  the  equipment  has  depreciated  to  zero, 
it,  together  with  the  accumulated  interest  thereon,   should 
amount  to  the  original  cost  of  the  equipment,  less  whatever 
scrap  value  it  may  have.    So  also  if  the  equipment  appre- 
ciates in  value,  this  amount  must  be  deducted  from  the  pro- 
duction cost  for  the  very   same   reasons   that   depreciation 
must  be  added.     The  Depreciation  Reserve  is  not  a  dona- 
tion to  the  Servant  from  the  employer,  but  a  trust  fund  that 


LIFE  19 

must  all  be  spent  for  the  purposes  for  which  it  was  allowed  or 
returned. 

Primarily,  the  amount  of  the  depreciation  reserve  that  must 
be  annually  laid  by  depends  upon  the  life  of  the  structure  or 
equipment.  This  life  must  be  determined  from  experience  and 
is  now  fairly  well  known.  It  is  given  as  well  as  possible  in 

the  table  below. 

TABLE  i 

APPROXIMATE  USEFUL  LIFE  OF  STRUCTURES  AND  EQUIPMENTS 

DESCRIPTION             LiFE,Yrs.  DESCRIPTION            LIFE,YRS. 

ASH  CONVEYERS,  steam  jet. ...    50             ELEVATORS,  bucket 10 

BRIDGES,  concrete Permanent      FANS,  centrifugal 25 

"        steel 40              GRATES 10 

"        wood 15             GENERATORS,  A.C 30 

BUILDING,  concrete Permanent                "            D.C 20 

brick. 50  HEATERS,  feed  water,  closed.  . .  20 

wood  or  sheet-iron. .    15                     "            "        "     open 30 

BOILERS,  fire-tube 15             HOISTING  MACHINERY 20 

"        water-tube 40  MOTORS same  as  generators 

BELTS,  leather 8              PUMPS,  plunger 15 

BINS,  steel 25                  "       centrifugal 25 

"     wood 10              PRODUCERS,  gas 20 

CHIMNEYS,  concrete Permanent      PLPING 35 

brick 50             ROTARY  CONVERTERS 30 

steel,  self-sustaining  35              STORAGE  BATTERIES 4 

sheet-iron 8              SWITCHBOARD 50 

CONDENSERS,  jet 20             TRANSFORMERS 50 

surface 20             TRANSMISSION  LINE,  pole 12 

CONVEYERS,  bucket 20                        "                "     steel 20 

belt 7             TRAVELING  CRANES 50 

ENGINES,  high-speed 15              TURBINES,  steam 20 

"         low-speed 25                      "          water 25 

ECONOMIZERS 15             WIRING,  electric 30 

The  above  list  is  necessarily  very  limited.  Besides  it  must 
be  borne  in  mind  that  the  life  given  is  necessarily  rather  approxi- 
mate. It  depends  greatly  on  the  quality  of  the  apparatus 
when  purchased,  together  with  a  multitude  of  other  conditions. 
The  above  table  is  for  good  substantial  equipment,  not  the 
cheap  stuff  that  is  also  found  on  the  market. 

Other  conditions  that  affect  life  are  the  surrounding  ele- 
ments. Thus,  the  life  of  wood  pipe  is  very  great,  if  it  is  kept 
thoroughly  impregnated  with  water,  but  deteriorates  rapidly 
with  disuse.  So  again  soil  conditions  often  are  a  deciding 
factor,  iron  pipe  decomposing  rapidly  in  alkaline  soil  or  sea 


20  COST  SEGREGATION 

water.  On  the  other  hand  the  life  of  stacks  is  greatly  influ- 
enced by  the  amount  of  sulphur  in  the  fuel.  This  is  due  to  the 
formation  of  sulphurous  acid  from  the  sulphur  dioxide  and  its 
oxidation  to  sulphuric  acid  in  the  presence  of  water  and  free 
oxygen,  when  the  temperature  is  not  too  high.  This  acid 
attacks  the  iron,  causing  very  active  corrosion.  It  seems  to  be 
most  active  during  light  or  no  load  periods.  So  also,  the  life 
of  boilers  being  fed  with  untreated  water,  depends  more  upon 
the  chemical  composition  of  the  raw  water  than  upon  any  one 
other  item. 

The  life  of  all  apparatus  depends  upon  the  skill  of  the  operat- 
ing engineers  as  well  as  the  proper  correlation  of  all  parts  of  the 
system  by  the  designing  engineer.  All  such  items  must  be 
taken  into  account  in  determining  the  probable  life  of  a  system. 
But  the  difference  between  normal  and  actual  conditions  should 
be  charged  to  management.  With  unusually  good  operators  a 
plant  will  tend  to  appreciate  in  value  for  quite  a  few  years. 
This  should  be  credited  not  only  to  the  operators  but  to  the 
management  which  had  the  courage  and  foresight  to  pay  the 
price  for  first  class  men.  On  the  other  hand  if  the  first  cost  of  a 
system  or  the  operating  cost  is  abnormally  high  due  to  poor 
designing,  the  difference  should  be  charged  to  bad  management 
where  it  ultimately  belongs. 

The  reader  is  cautioned  not  to  confuse  depreciation  with 
wear,  as  is  often  done  in  practice.  Depreciation  is  a  fixed 
charge,  continuing  unaltered  whether  the  unit  is  in  use  or 
idle.  Wear  is  an  operating  cost,  the  amount  of  wear  being  in 
direct  proportion  to  the  amount  of  use.  When  a  machine  is 
worn  out,  it  has  not  depreciated  to  zero  worth.  It  has  merely 
worn  out  and  we  have  merely  a  major  repair  item. 

Thus  when  a  bearing  is  worn  out,  we  replace  the  bearing  and 
thus  repair  the  engine.  This  is  admittedly  a  repair  item. 
So  when  an  engine  is  worn  out,  we  replace  the  engine  and  thus 
repair  the  power  plant.  The  replacement  of  an  entire  worn- 
out  engine  is  just  as  much  a  repair  item  as  the  replacement  of 
any  of  its  parts,  the  only  distinction  being  the  degree  of  the 
replacement.  Whenever  the  cost  of  maintenance  of  an  old 


MATERIALS  CONSUMED  21 

unit  is  greater  than  the  total  costs  (fixed  charges  and  mainte- 
nance) of  a  new  unit,  it  is  evident  that  it  will  pay  us  to  discard 
the  old  unit.  In  such  a  case  it  is  worn  out. 

We  may  therefore  distinguish  two  lives  for  all  things,  namely 
the  natural  life  and  the  operating  life.  The  natural  life 
depends  on  the  factors  of  obsolescence,  inadequacy  and 
uselessness.  The  operating  life  depends  upon  the  amount  of 
use  that  may  be  obtained  from  anything  before  it  is  worn 
out.  The  shorter  of  the  two  determines  the  true  life  of  the 
unit. 

It  would  be  senseless  and  very  bad  engineering  to  make  the 
operating  life  of  a  unit  longer  than  its  natural  life.  Yet  this 
very  thing  has  been  done  in  the  past.  The  large,  extremely 
low  speed  engines  of  forty  and  fifty  years  ago,  some  of  which 
are  still  in  use,  refuse  to  wear  out,  though  they  are  hopelessly 
obsolete.  It  is  for  this  reason  that  practically  none  of  this 
type  of  machinery  is  now  built. 

26.  Taxes  may  be  definitely  determined  in  any  community, 
usually  running  between  one  and  two  per  cent  of  the  assessed 
value.     The  assessed  value  is  usually  somewhat  below  the 
original  cost  of  the  undertaking  and  this  must  of  course  be 
taken  into  account. 

Insurance  usually  runs  between  0.5  and  1.5%  but  this  again 
depends  upon  the  construction  of  the  system.  A  concrete 
building  that  is  absolutely  fireproof  need  carry  no  insurance, 
thus  tending  to  reduce  the  total  cost  of  such  a  structure.  If, 
for  example,  a  building  cost  $100,000  and  insurance  is  i%, 
then  the  annual  insurance  outlay  is  $1000  which,  at  5%, 
interest,  capitalizes  at  $20,000.  A  fireproof  building  would  be 
worth  this  much  more.  But  to  all  this  we  must  add  the  cost 
of  insurance  on  the  equipment  that  is  to  be  housed  within 
the  building,  further  increasing  the  value  of  the  firepoof 
structure. 

27.  Materials  Consumed.  —  Any  undertaking  rendering  a 
given  service,  whether  a  factory  producing  a  given  commodity 
or  a  power  plant,  uses  up  a  certain  amount  of  materials,  which 
is  either  entirely  consumed  or  else  rendered  into  a  more  or  less 


22  COST  SEGREGATION 

worthless  condition.  The  amount  of  material  so  consumed  is 
very  closely  in  direct  proportion  to  the  amount  of  service 
produced. 

Thus  a  steam  power  .plant  consumes  fuel,  lubricating  oils  and 
greases,  waste,  packing,  paints,  and  the  like.  A  hydro-electric 
power  plant  consumes  all  of  these  except  fuel.  A  factory  pro- 
ducing machinery  consumes  a  great  variety  of  supplies  as 
well  as  iron,  steel,  brass,  and  the  like  in  considerable  quanti- 
ties, which  are  rendered  into  the  more  or  less  useless  form  of 
turnings  and  punchings. 

The  cost  of  all  such  materials  consumed  should  be  charged 
under  this  head  to  operation,  together  with  their  cost  of  pur- 
chase and  delivery,  and  the  cost  of  waste  removal,  less  the 
salvage  if  any. 

28.  Attendance.  —  The  salary  or  wages  paid  to  all  who  are 
actually  engaged  in  the  production  of  a  given  service  should 
be  charged  to  operation  under  the  head  of  attendance.    In  an 
iron  works,  for  example,  this  includes  the  wages  paid  to  machin- 
ists, molders,  boiler  makers,  helpers,  clerks,  bookkeepers,  col- 
lectors, and  the  like,  together  with  such  parts  of  the  salary  of 
the  superintendent,  engineer,  and  manager  as  are  devoted  to 
the  actual  production  of  the  service.    And  so  for  other  under- 
takings. 

29.  Maintenance. -- The  cost  of  all  materials  and   labor, 
engineering,  and  inspection  service  necessary  to  keep  an  under- 
taking in  good  running  condition,  so  that  its  productive  capacity 
does  not  decline,  should  be  charged  to  operation  under  the 
head  of  maintenance.    The  largest  item  that  comes  under  this 
head  is  repair  and  replacement  of  worn-out  parts  of  equipment 
and  the  maintenance  of  grounds  and  buildings.     But  as  pre- 
viously pointed  out,  the  cost  of  equipment  replaced  because  of 
obsolescence,  inadequacy,  or  uselessness  should  be  assessd  to 
fixed  charges  under  the  head  of  depreciation.     Attendance, 
maintenance,    and    materials  consumed    decrease    with    the 
increase  in  quality  of  the  system,  while  its  first  cost  increases 
therewith.     Therefore  as   the   fixed   charges   are   allowed   to 
increase,  the  operating  costs  decrease.     This  gives  us  one  of 


STORAGE,  SALE  AND  DISTRIBUTION  23 

our  most  important  problems,  to  find,  under  given  load  con- 
ditions, where  the  sum  of  the  fixed  charges  and  operating 
costs,  i.e.,  the  total  costs  of  production  of  the  service,  are  a 
minimum. 

30.  Stock   Account.  —  It   is   invariably  necessary   in   any 
undertaking,  whether  large  or  small,  to  carry  a  special,  tempo- 
rary account  of  materials  on  hand  for  use  or  sale.    This  account 
should  be  divided  into  two  parts,  namely  raw  .stock  and  fin- 
ished product.     Thus  in  a  steam-electric  power  plant,  we  have 
on  hand  usually  considerable  materials,  such  as  fuel,  supplies, 
and  the  like,  which  should  be  charged  to  capital  until  used  and 
then  transferred  to  operation.     In  this  case  we  have  no  fin- 
ished product,  for  electricity  is  and  must  be  used  as  produced, 
excepting   what   little   is    carried  over  in  storage  batteries. 
In  the  case  of  an  engine  factory,  we  have  the  raw  materials 
account,  covering  materials  consumed  as  well  as  those  going 
into  the  finished  product.    This  stands  as  a  debit  against  the 
production  of  the  engines,  while  the  finished  product  stands 
as  a  credit  account. 

31.  Storage,  Sale  and  Distribution.  —  As  previously  shown, 
the  costs  of  rendering  a  given  service  may  be  divided  into  col- 
lection, production,  transportation,  and  distribution,  the  latter 
including  sales.    The  item  of  storage,  sales,  and  distribution, 
while   in  some  business  the  most  important  function,  is  no 
exception  to  the  treatment  given  any  of  the  other  items  in  the 
complete  rendition  of  a  service.      The  more  uniformly  we 
operate  throughout  the  year  the  more  must  be  stored,  i.e.,  we 
can  decrease  the  production  cost  at  the  expense  of  the  storage 
cost.    To  determine  the  conditions  of  best  economy,  in  such  a 
case,  gives  a  typical  problem  in  cost  analysis. 

Selling  is  one  of  the  most  important  services  that  is  rendered, 
because  it  comprises  not  only  the  actual  selling  but  the  giving 
of  advice  and  assistance  to  the  purchaser  in  the  successful  use 
of  the  product  being  sold,  as  well  as  trouble  shooting,  that 
is  the  correction  of  troubles  that  appear  in  the  use  of  the  prod- 
uct whether  due  to  more  or  less  defect  in  the  product  or  to 
difficulties  that  the  purchaser  gets  himself  into. 


24      ^  COST   SEGREGATION 

32.  In  any  actual  plant  all  the  above  items  of  expense  are 
found  to  be  more  or  less  mixed  together  since  operation,  main- 
tenance, replacement,  and  the  like  are  taking  place  simul- 
taneously. They  can,  of  course,  be  readily  segregated  by  an 
engineer,  not  with  absolute  accuracy,  but  sufficiently  so  for  the 
most  exacting  practical  needs. 


CHAPTER   III 
FUNDAMENTAL  FINANCIAL  CALCULATIONS 

Compound  Interest.  Principal.  Equity.  Term  Factor.  Operating 
Vestance.  Depreciation  Rate.  Present  Worth  of  a  Depreciating 
Equipment.  Depreciation  Vestance.  Total  Vestance.  Annual 
Operating  Costs. 

33.  Interest.  —  The  loan  of  money  is  paid  for  by  the  annual 
payment  of  interest  usually  expressed  as  a  certain  per  cent  of 
the  principal.    The  loan  of  money  is  always  for  a  certain  term 
of  years  only,  at  the  end  of  which  time  the  principal  must  be 
returned,  the  interest  in  the  meantime  being  paid  annually. 
This  is  so-called  Simple  Interest.    On  the  other  hand,  instead 
of  paying  the  interest  when  due,  it  may  be  added  to  the  prin- 
cipal and  become  a  part  thereof,  the  sum  of  the  two  thereafter 
bearing  interest.    This  is  called  Compound  Interest.    Thus  we 
meet  the  problems  of  how  much  would  a  principal  amount  to 
in  a  certain  number  of  years,  bearing  compound  interest  at  a 
given  rate,  and  conversely,  knowing  that  a  certain  amount  is 
due  in  a  given  number  of  years  at  a  known  rate,  to  find  its 
present  value. 

34.  Compound  Interest  Formula.  —  Letting 
P  =  the  principal, 

R  =  the  rate  of  interest, 
N  =  the  number  of  years, 
and     A  =  the  amount  at  compound  interest, 

then  A  =  P(i  +  /?) 


« 


This  can  be  easily  shown  as  follows  : 
Example  i.   To  begin  with  we  have 
AQ  =  P  dollars. 

25 


26         FUNDAMENTAL  FINANCIAL  CALCULATIONS 

At  the  end  of  the  first  year,  this  has  increased  to 

Al  =  P(i  +  R),  • 
so  at  the  end  of  the  second  year,  it  is 

A,  =P(i  +K)  (i  +  R)  =P(i+R)\ 
And  at  the  end  of  the  third  year,  we  have 

A3  =  P(i  +  R)*  (i  +  R)  =  P(i  +  tf)'3, 
and  so  on.    At  the  end  of  (n)  years,  it  is  evidently 

An    =  P(l    +R)" (i) 

Example  2.   How  much  will  $100  amount  to  in  10  years  at 
8%  compound  interest? 

Solution:  In  this  case 

A  =  ioo(i.o8)10 
or  log  A  =  log  100  -f  10  log  1.08 

=    2    +   10    X   0.0334    =    2.334, 

so  that  A  =  $215.80. 

Example  3 .    In  how  many  years  will  a  sum  of  money  treble 
itself  at  10%  compound  interest? 
Solution:   In  this  case  we  have 


where 


p  ~3 


and  R  =  o.io, 

so  that  3  =  (i.i)n, 

and  n  log  i.i    =  log  3, 

Iog3         Q-477I 

or  n  =  ; =  , 

log  i.i       0.0414 

whence  n  =  11.52  years. 

35.  Principal.  —  Knowing  the  sum  of  money  that  will  be 
due  in  a  certain  number  of  years  at  a  given  rate  of  compound 
interest,  we  can  find  its  present  worth,  by  solving  the  above 
formula  for  (P),  thus 

P  =  -  -^-  (2) 

' 


EQUITY  27 

Example  4.    One  thousand  dollars  is  due  in  ten  years,  the 
compound  interest  rate  being  7%.    What  is  its  present  worth? 

Solution:  In  this  case 

A  =  $1000, 

R  =  7% 
and  n  =  10, 

„         1000 
so  that  P  =  -, TTT, 


or  log  P  =  log  1000  —  10  log  1.07 

=  3  -  10  X  0.0294  =  2.706, 
whence  P  =  $508.10. 

36.  Equity.  —  If  the  operation  of  a  certain  equipment  is 
guaranteed,  and  its  actual  performance  falls  below  this,  the 
cost  of  operation  of  this  unit  will  be  increased  by  a  certain 
amount  annually.  In  such  a  case  a  certain  sum  (the  equity) 
must  be  deducted  from  the  price  of  the  unit  to  compensate  the 
purchaser  for  this  loss  in  guaranteed  economy.  On  the  other 
hand,  if  the  guarantee  is  exceeded,  a  certain  sum  should  be 
added  to  the  price  to  compensate  the  manufacturer  for  the 
increased  economy  attained. 

In  determining  the  equity,  it  is  evident  that  if  we  capitalize 
this  annual  amount  by  which  the  guaranteed  economy  falls 
short  (or  is  exceeded)  at  the  usual  rate  of  interest,  and  pay 
this  sum  to  the  purchaser,  he  will  receive  this  annuity,  as 
interest  therefrom  forever,  whereas  the  equipment  has  a 
limited  life  only.  We  would  thus  be  paying  too  much.  Call- 
ing (a)  the  annuity,  or  the  annual  cost  of  operation  in  excess 
of  the  guarantee,  and  (R)  the  interest  rate,  then  this  capitalized 
amount  (A)  will  equal  merely 


Instead  of  deducting  (A)  dollars  from  the  selling  price  (C) 
which,  as  pointed  out  above,  would  be  too  large  a  sum,  we 
should  loan  this  sum  (A)  to  the  purchaser  during  the  life  of  the 
unit,  without  charge,  the  principal  to  be  returned  by  him  at  the 
end  of  this  time,  the  interest  thereon,  would  compensate  the 


28         FUNDAMENTAL  FINANCIAL  CALCULATIONS 

purchaser  for  the  reduced  efficiency.  This  would  be  an  equi- 
table adjustment.  But  this  adjustment  can  be  simplified.  If 
instead  of  waiting  for  (n)  years,  the  life  of  the  equipment  before 
the  amount  (^4)  is  returned,  we  deduct  from  this  amount  (A), 
due  in  (n)  years,  its  present  worth  (P),  then  we  can  reach  a 
complete  adjustment  at  once.  This  difference  (A  -  P)  is  the 
equity  (£),  so  that 

E  =  (A  -  P), 
but  '-(Tng;, 

and  A  =  ~» 


37.  Alternate  Solution.  —  We  can  obtain  the  above  formula 
in  a  more  direct  but  also  more  laborious  way  as  follows:  To 
begin  with  we  lay  aside  E  dollars.  At  the  end  of  the  first  year, 
this  has  increased  to 

E  (i  +  R)  dollars, 

from  which  we  must  pay  the  purchaser   (a)  dollars.     So  we 

have  left 

[E(i  +  R)-  a]  dollars. 

This  will  increase  to 

[£(i  +  R)  -  a]  (i  +  R)  =  E(i  +  R)2  -  a(i  +  R)  dollars 

at  the  end  of  the  second  year,  when  we  must  pay  out  the 
second  installment  of  the  annuity.  After  this  we  have  left 

E(i  +  R)2  -  a(i  +  R)  -  a 

which  will  increase  to  [E(i  +  #)3  -  a(i  +  R)2  -  a(i  +  P)] 
at  end  of  the  third  year.  So  then  at  the  end  of  the  third 
year,  we  have  left  after  paying  the  third  installment  of  (a) 
dollars 

E(i  +  #)3  -  a(i  +  R)2  -  a(i  +  R)  -  a, 


EQUITY  29 

and  so  at  the  end  of  (n)  years  we  have  left 

E(i  +  RY  -  a(i  +  RY'1  -  a(i  +  ^)n~2  .    .    . 

a(i  +  R)2  -  a(i  +  R)  -  a. 

If  (n)  is  the  life  of  the  equipment  —  the  term  of  the  annuity 
-  then  the  balance  left  on  hand  should  be  zero,  so  that 


-  a(i  +  R)  -  a~]  =  o, 
or 

E(i  +  RY  -  fl[(i  +  #)n-!  +-(i  +  ^)n~2  •    •    • 

+  (i  +  J?)  +  i]. 

We  have  next  to  find  the  sum  of  the  series  on  the  right  hand. 
Calling  this  sum  (S)  ,  then 

£(i  +  RY  =  aSy 

and  S  =  (i  +  RY-1  +  (i  +  RY~2  '    '    '    +  (i  +  R)  +  i. 
Multiplying  through  by  (i  +  R),  we  get 
(i  +  R)S  =  (i  +  #)"  +  (i  +  tf)"-1  •    .    . 

+  (i  +  £)3  +  (i  +  £)2  +.(i  +  ^)- 

Now  subtracting  (5)  from  the  left  hand  side  of  this  equation,  and 
its  value,  the  foregoing  series,  from  the  right  hand  side,  we  get 

(i  +  K)S  -  S  =  (i  +  RY  -  i, 
or  RS  =  (i  +  RY  -  i, 

(i  +  RY  -  i 
and  o  =  -      —  ^  — 

But  since  E(i  +  i?)n  =  aS  we  can  substitute  for  (S)  its  value 
just  found  and  get 


or 

«  f(i  +  *>  -  il      a  f 

Ri     (i+RY     ]      Rl        (i 
as  before. 

38.  Example  5.  A  100  h.p.  engine  is  guaranteed  to  consume 
not  over  2O#  of  steam  per  h.p.h.  A  test  shows  its  actual  con- 
sumption to  be  22#.  If  the  engine  is  run  at  full  load  for  3000 


30        FUNDAMENTAL  FINANCIAL  CALCULATIONS 

hours  per  year,  and  the  steam  costs  20  cents  per  iooo#,  what 
will  be  the  amount  which  should  be  deducted  from  the  pur- 
chase price  of  the  engine  to  compensate  for  this  loss  in  econ- 
omy? The  life  of  the  engine  is  20  years;  interest  rate  5  %. 

Solution:  Under  the  above  conditions,  the  engine  evidently 
uses  2oo#  of  steam  per  hour  in  excess  of  the  guarantee.  In  a 
year  of  3000  hours  it  would  use  in  excess 

3000  X  200  =  6oo,ooo#, 

..  600,000  ft 

costing  - X  0.20  =  $120, 

1000 

so  we  have  that  a  =  $120, 

R  =  0.05, 

n  =  20, 


whence 


^       1 20     (i.cx)20  —  i 

E  -  —-         •  24°° x 


or  E  =  $1239. 

39.  Example  6.  A  300  h.p.  motor  is  sold  for  $1800  with  a 
guaranteed  efficiency  of  92  %.  It  is  used  at  full  load  for  3000 
hours  per  year.  The  life  of  the  motor  is  20  years  and  the  power 
costs  one  cent  per  h.p.h.  If  the  motor  only  develops  90%  effi- 
ciency, how  much  should  be  deducted  from  the  price  to  com- 
pensate the  purchaser  therefor,  assuming  an  interest  rate  of 


Solution:  The  horse  power  that  would  be  consumed  by  the 
motor  under  the  guarantee  is 

300  -5-  0.92  =  326  h.p., 
while  it  actually  consumes 

300  -5-  0.90  =  333.3  h.p., 
a  difference  of 

333-3  -  326  =  7-3  h.p., 

costing  7.3  cents  per  hour  or 

0.073  X  3000  =  $219  per  year. 


TERM  FACTOR  31 

So  we  have  here  that 

a  =  $219, 
R  =  0.06, 
n  =  20, 
so  that 

219    (i.o6)20-i 
"  0.06'      (i.o6)20 

or 

E  =  $2450. 

The  corrected  price  would  then  be 

1800  —  2450  =  —  $650. 

That  is,  assuming  we  can  get  a  motor  of  92  %  efficiency  from 
another  firm,  we  could  not  afford  to  accept  this  motor  as  a 
present,  unless  we  received  therewith  a  cash  bonus  of  $650. 

40.   Term  Factor.  —  The  factor  ^  +  f^~  T  we  shall  desig- 

(i  +  R)n 

nate  for  convenience  the  term  factor  (T),  so  that 

aT  (  N 

...  £=¥ '  '  '   (4) 

In  the  table  below,  we  give  the  values  of  this  term  factor  (T) 
for  various  values  of  the  interest  (R)  and  the  life  (n). 


FUNDAMENTAL  FINANCIAL  CALCULATIONS 


II 

p 


ONOOM 
O\    O     l- 

O    ON  to 


to  O    co  to 


O\M 

M     PO   J>-  O     Tj-    ON  OO     t^CN 
O    M    t^.   (N     ^  J>.   •rf  O    ON 


O     -<TO     J>-    ON 


O  ,M      M      M    OO      O       ON 

t^cs     M    t^  r>.   co  O     co  t^  OO  O  O     <s     toO     M     O    t^   co   <N  csO    to   M     TtM 

to   co  OO    to   O    f^   ON   M    to  to   Tt~    Tj~  OO    t^*   ^*    M     O     ^O   O  O  O     M    to  co   ^  OO 

OO     tO«N     ONtoO     toONCOf^M     Tj-t^.O     CN     TfoO     <NOO     <N  toOOO     ONONON 


t^.vO'O 

O^O     M 


oN 

(N 


<N\O  100 

t^Tj-o\ 
ONiOfO 
rj-oo  io 


<N\O     O     w     (SO    toOOO    lOOO 
rt-tot^M     OO     cs     ONQ'OOO 
to  vO     ^  i>*  t^.   fo  ^*  t^»  vO     HH     ^   to 
vO     <NOO    fOOO     rot^M    IOO\<N    tooo     M 
O     M    M    <N     <N     cofO-^-Tj-Tt-ioiOiOVO 


t^.r}-OO  lO 
cs  M  to  t^-  " 
w  lo 


vO 
O 


O     O 

M\O 


toOO  tOH  OOOO 
i^rOOO  CN  QOO  O 
MvOOOOO  t^toOs 
oO  O  ^00*0  <N^O 


. 

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O     O 


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M  ^O  COM 
fOto\O  torO 
CN  tooo  H  rJ- 
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M'OO  t^O\t^^<N  ro 
cOTfoOOO  t^-ovO  ^CN 
CN  OO  M  IOM  •^•t^ON 
'O  t^t^OOOO  ON<^ONO\ 


M      T+- 
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T^-^-Q     M     M 


O    O 

o 


MVO       MOO      t^OO      MOO      ^O      O 

M  ("O  vO  OO^O  tot^tN  O  t^-M 
^OvO  rO'cJ-MVO  M  Ostot^O 
O  O  rt-<N  O\^c\toO  ^oo 


<N     Tl-toM     MOO     rO<N     Q     ONO   00     CO    ON  O     «*5    ON   <N     M     rf    M     ON    O\  O  »^   'T 

M     -400    lO^-Tj-ONVO'OOO    toO     OOO     MOO    toro-^-OO     <NOO     <N  O     O 

Q\t^TtM     t^.<NVO      O     PO^Ot^OO     ONOOOOO     MO     <NOO     rl-tOM     O  MOO 
M    1000     M     roOOO     M     rotot^O\M     ro 


roOOO     M     rotot^O\ 
MMMCSMWW« 


M    r*-(NOO 


»oO    too    O    O    ^oQ^d 

WCOCO'TlOOl^O.M 
M    ^^ 


DIFFERENCE  IN  WORTH  33 

Example  7.  What  is  the  difference  in  worth  per  h.p.  of  two 
Diesel  engines,  one  of  which  uses  o.4o#  of  oil  per  h.p.  and  the 
other  uses  0.46$  oil  per  h.p.  if  the  engines  are  used  at  full  load 
for  4000  hours  per  year,  the  oil  costing  0.5  cent  per  pound, 
interest  7  %  and  the  life  of  the  engines  being  25  years? 

Solution:  The  difference  in  fuel  used  per  h.p.h.  is  o.o6#  and 
per  year  is  0.06  X  4000  =  240$,  costing  $1.20.  So  we  have 

a  =  $1.20, 

R  =  0.07, 

n  =  25  years. 
Whence  from  table  2, 

T  =  0.81575 

and  E  =   -1—   X  0.81575, 

0.07 

or  E  =  $13.9843. 

That  is  the  engine  using  only  0.40$  oil  per  h.p.h.  is  worth 
about  fourteen  dollars  more  per  h.p.  than  the  engine  using  0.46$. 
Evidently  it  would  not  pay  to  spend  more  than  -this  amount 
to  gain  this  added  economy. 

Example  8.  What  is  the  difference  in  worth  between  a 
steam  plant  using  2#  of  coal  per  h.p.h.  and  a  producer  plant 
using  i#  coal  per  h.p.h.,  the  coal  costing  $5  per  ton  and  interest 
being  6%,  if  the  lives  of  the  plants  are  20  years,  and  they  are 
run  at  full  load  for  3000  hours  per  year,  other  things  being  equal? 

Solution:  At  $5  per  ton,  the  coal  costs  J  cent  per  pound. 
The  steam  plant  therefor  cost  J  cent  per  h.p.h.  more  than  the 
producer  plant,  or  J  X  3000  =  $7.50  more  per  year  per  h.p. 
So  then  we  have 

a  =  $7.50, 

R  =  6%, 
n  =  20  years. 
Whence  by  table  2, 

T  =  0.68821, 

and  E  =  ^^  x  0.68821, 

0.06 

or  E  =  $86.02625. 


34        FUNDAMENTAL  FINANCIAL  CALCULATIONS 

That  is,  under  the  conditions  assumed,  the  producer  plant  is 
Worth  86  dollars  more  per  h.p.  than  the  steam  plant. 

Example  8.  Determine  how  much  more  we  could  pay  for  a 
hydro-electric  power  plant  and  transmission  line,  than  for 
a  steam  plant,  to  be  run  at  full  load  for  3000  hours  per  year, 
if  the  operating  costs  for  the  latter  are  0.6  cent  per  h.p.h., 
while  for  the  hydro-electric  system,  they  are  only  0.2  cent. 
Assume  the  life  in  either  case  to  be  30  years  and  interest  to  be 
5  %  while  all  other  things  are  equal. 

Solution:  In  this  case  it  costs  0.4  cent  more  per  h.p.h. 
for  the  steam  plant  than  for  the  hydro-electric  plant,  or 
0.004  X  3000  =  $12.  more  per  h.p.  year.  So  we  have 

a  —  12.00, 

R  =  5%, 
n  =  30. 
whence  by  table  2, 

T  =  0.76861, 

and  E  =  i^  X  0.76861, 

0.65 

or  E  —  $184.4664. 

Example  9.  A  Diesel  engine  plant  uses  0.4$  of  fuel  oil  per 
h.p.h.  costing  $1.60  per  barrel  (320#),  while  the  attendance 
cost  is  o.i  cent  per  h.p.h.  A  steam  plant  uses  3$  coal  per 
h.p.h.  costing  $5  per  ton  with  an  attendance  cost  of  0.2  cent 
per  h.p.h.  These  plants  are  assumed  to  have  a  life  of  15  years. 
If  interest  is  5  %  and  the  plants  are  to  be  run  for  4000  hours 
per  year  at  full  load,  what  is  the  difference  in  worth  between 
the  two? 

Solution:  At  $1.60  per  barrel,  the  oil  costs  0.5  cent  per  lb., 
so  that  for  the  Diesel  Engine  Plant  we  have 

Fuel  cost  =  0.4  X  0.5  =  0.2  cent, 
Attendance  =  o.i  cent, 
Operating  cost  =  0.2  -f-  o.i  =  0.3  cent. 


OPERATING  VESTANCE  35 

At  $5  per  ton,  the  coal  costs  0.25  cent  per  lb.,  so  that  for 
the  steam  plant  we  have 

Fuel  cost  =  0.3  X  0.25  =  0.75  cent, 
Attendance  =  0.2  cent, 
Operating  cost  =  0.75  +0.2  =  0.95  cent. 

Evidently  the  steam  plant  costs  0.95  —  0.3  =  0.65  cent  more 
per  h.p.h.  than  the  Diesel  plant,  or  0.0065  X  4°°°  =  $26  more 
per  year.  So  we  have 

a  =  $26, 

n  =  15  years, 

and  R  =  5%. 

Whence  by  table  2, 

T  =  0.51897, 

tyf\   f\C} 

and  E  =  -      -  X  0.51897, 

0.05 

or  E  =  $269.8644. 

41.  We  have  thus,  as  illustrated  in  the  above  problems,  a 
method  not  only  for  determining  the  equity  in  any  given  case, 
but  the  same  method  permits  us  to  get  the  difference  in  value 
of  various  units  or  systems  only,  however,  under  full  load 
conditions.  The  case  of  relative  values,  and  equities  under 
partial  load  conditions,  will  be  taken  up  later.  We  can  extend 
this  treatment  somewhat  further  by  determining  the  capital- 
ized value  of  the  operating  cost.  Thus  if  the  operating  cost 
is  (a)  dollars  per  year,  and  since  this  outlay  will  continue  for  all 
times,  and  not  merely  for  the  life  of  the  equipment,  its  total 
capitalized  amount,  or  operating  vestance  (  V)  as  we  shall  call 
it,  will  be 


where  (R)  is  the  rate  of  interest  as  before  and  (a)  the  annual 
operating  cost. 

Example  10.  If  it  costs  $12  per  year  for  operation  to  produce 
a  continuous  horse  power  and  money  costs  us  6  %,  what  is  the 
operating  vestance  (V)  per  h.p.? 


36         FUNDAMENTAL  FINANCIAL  CALCULATIONS 

Solution:  Here 

a  =  $12, 
R  =  6%, 

so  that  V  =  —  =  $200. 

0.06 

That  is  by  investing  $200  at  6  %,  the  proceeds  thereof  will  pay 
the  operating  cost. 

42.  The  Depreciation  Rate  is  a  certain  per  cent  of  the  first 
cost  which  is  annually  laid  aside  to  form  the  depreciation 
reserve.    This  rate  must  be  such  that  at  the  end  of  the  life 
of  the  equipment,  the  depreciation  reserve,  together  with  its 
accumulated  interest,  will  equal  the  first  cost  less  the  scrap 
value  of  the  equipment.     The  item  of  accumulated  interest 
on  the  depreciation  reserve  has  often  been  overlooked  in  the 
past,  introducing  a  very  great  error.    Thus  if  the  life  of  a  cer- 
tain item  of  equipment  is  50  years,  it  has  usually  been  assumed 
that  the  depreciation  rate  was  2  %.    We  shall  show  later  that 
this  is  about  six  times  too  large  —  an  error  of  600%. 

43.  To   solve   this  problem  correctly,  we  must  determine 
an  amount  or  annuity  (^4)  which,  when  laid  aside  each  year 
for  (n)  years,  the  life  of  the  equipment,  and  drawing  interest 
at  (R)  per  cent,  compounded  annually,  will  amount  to  (W) 
dollars,  the  first  cost  of  the  unit  less  its  scrap  value,  i.e.,  its 
wearing  value.     The  depreciation  rate  (D)  which  should  be 
applied  only  to  the  wearing  value  (Traction  Valuation  Com- 
mission of  Chicago,  1906)  is  then  merely, 


It  is  difficult  to  determine  for  long  periods  in  advance  the  scrap 
value  of  any  item  of  a  system,  and,  furthermore,  this  item  is 
small  in  comparison  with  the  first  cost.  In  practice  it  is  cus- 
tomary to  neglect  this  difference  and  apply  the  depreciation 
rate  to  the  first  cost  (C),  getting  the  approximate  equation 


DEPRECIATION  37 

This  will  be  correct  if  it  is  understood  that  at  the  time  of  a 
purchase  of  a  new  unit  the  scrap  value  of  the  replaced  unit  is 
applied  to  this  purchase  and  not  treated  as  an  asset  of  the  un- 
dertaking. This  will  give,  then,  exact  results,  when  an  attempt 
to  determine  the  scrap  value  of  a  unit,  15,  20,  or  30  years 
hence  would  be  only  a  wild  guess.  The  determination  of  the 
depreciation  rate  follows,  as  per  these  latter  conditions. 

44.   Let 

A  =  the  depreciation  annuity  (or  amount), 

R  =  the  interest  rate, 
C  =   the  first  cost, 
n  =   the  life  of  the  unit, 
and  D  =   the  depreciation  rate. 

Then  at  the  end  of  the  first  year,  we  deposit  (^4)  dollars,  so 
that  the  depreciation  reserve  is 

FI  =  A  dollars. 

At  the  end  of  the  second  year,  due  to  the  interest  rate,  this  will 
increase  to  A(i  +  R)  dollars,  to  which  we  add,  by  deposit, 
another  (A)  dollars.  The  depreciation  reserve  is  then 


F2  =  [A(i  +  R)  +  A~]  =  4[(i  +  R)  +  i]  dollars. 

At  the  end  of  the  third  year  this  increases  to  [>4(i  +  R)  +  A~] 
(i  +  R),  to  which  we  add  another  (A)  dollars  by  deposit, 
making  the  reserve  then 

F3  =  A[(i  +  R)  +  i]  (i  +  R)  +  A 
or  F,  =  A[(i  +  R)*  +  (i  +  R)   +  i], 

so  at  the  end  of  the  fourth  year  it  is 

Ft  =  A[(i  +  tf)3  +  (i  +  R)2  +  (i+R)  +  i], 
and  at  the  end  of  the  (n)  year,  the  depreciation  reserve  is 

Fn  =  A[(i  +  R)"-1  +  (i  +  R)-2  .    .    . 

H-  (i  +  R)2  +  (i  +  R)  +  i]  dollars. 


38        FUNDAMENTAL  FINANCIAL  CALCULATIONS 

Again,  calling 

S  =  (i  +  R)*-1  +  (i  +  R)n~*  .    .    .    +  (i  +  *)»  +  i, 
then 

(i    +*)»-! 

R 
so  that 


But  at  the  end  of  (w)  years,  where  (w)  is  the.  life  of  the  equip- 
ment, then  the  depreciation  reserve  (Fn)  must  equal  (C),  the 
first  cost,  whence 

c  =  |[d  +  *)•  -  i], 

or 

A  R 


C      (i  +  R)»  -  i 
But  the  depreciation  rate  (D)  equals  — ,  by  eq.  (7),  so  that 

C_x 

D_          R  .  ,•<*          (8) 

Example  n.  The  life  of  an  engine  is  10  years,  interest  rate 
6%.  What  is  the  depreciation  rate? 

Solution:  Here  n  =  10  years, 
R  =  0.06, 

so  that  D  =  -—  -  =  7.5%, 

(i.o6)lu  —  i 

instead  of  10%  usually  assumed. 

Example  12.  If  in  example  n,  the  interest  rate  is  8% 
instead  of  6%,  what  will  be  the  depreciation  rate? 

Solution:  Then  n  =  10  years, 
R  =  8%, 

0.08 
whence  D  =  (i>o8)10  _  T  =  6.9%. 

The  larger  the  interest  rate  or  the  longer  the  life  of  the 
equipment,  the  smaller  the  depreciation  rate,  as  illustrated 
below. 


APPRECIATION  39 

Example  13.     The  life  of  a  concrete  building  is  50  years, 
interest  rate  6%.    What  is  the  depreciation  rate? 

Solution:  Here  R  =  6%, 

n  =  5°  years, 
0.06 


sothat       '       D  = 


0.06)°  - 


That  is,  the  depreciation  rate  is  approximately  one-  third  of 
one  per  cent  instead  of  the  2%  usually  assumed,  giving  an 
error  of  some  600%  as  mentioned  before. 

In  order  to  facilitate  calculations,  we  give  below,  in  Table  3, 
the  depreciation  rate  for  various  values  of  (n)  and  (R),  accord- 
ing to  equation  (8). 

45.  We  have  spoken  thus  far  only  of  depreciation. 
Machinery  depreciates  with  age.  But  lands,  live-stock,  and 
the  like  almost  invariably  appreciate  in  price.  Evidently 
appreciation  is  merely  negative  depreciation,  so  that  if  we 
charge  depreciation  as  a  cost,  then  we  must  charge  appreciation 
as  a  profit. 

Lands  invariably  increase  in  value  with  the  increase  in  popu- 
lation in  their  vicinity  or  in  the  development  of  their  resources 
previously  lying  undeveloped  and  perhaps  undiscovered.  The 
increase  in  price  of  land  with  population  is  very  pronounced. 
Live-stock  appreciates  by  natural  processes.  But  besides  all 
this,  all  classes  of  things  may  apparently  increase  in  price  due 
in  reality  to  a  depreciation  of  the  currency.  Under  such  condi- 
tions, they  really  do  not  increase  in  worth  at  all.  A  pound  of 
wheat  has  no  more  nutritive  value  now  than  a  hundred  years 
ago"  though  it  now  costs  many  times  as  much.  Its  nutritive 
value  is  a  constant.  Its  increased  cost  is  only  apparent  due 
to  the  decreased  price  of  the  money. 

So  a  piece  of  machinery  may  increase  in  price,  but  not  worth, 
due  likewise  to  the  decreased  price  of  gold.  Where  the  price 
of  the  metals  and  labor  has  doubled,  it  is  evident  that  the 
price  of  the  machine  must  be  doubled,  which  merely  means, 
according  to  our  present  system  of  exchange,  that  it  will  take 


FUNDAMENTAL  FINANCIAL  CALCULATIONS 


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APPRECIATION  41 

twice  as  much  gold  to  buy  the  same  machine  of  the  same  worth 
as  before. 

The  increased  production  of  gold  results  in  a  decrease  in  its 
price.  The  price  is  determined  by  the  demand  for  the  gold  for 
such  useless  uses  as  ornamentation  and  exchange  and  for 
useful  use  in  industry.  The  price  that  can  be  paid  for  gold  for 
industrial  use  is  strictly  limited  by  its  intrinsic  worth.  The 
price  it  may  have  for  useless  uses  is  a  problem  of  mob  psy- 
chology rather  than  one  of  science.  As  the  production  increases, 
the  price  falls,  and  with  the  fall  in  price,  more  gold  is  used  in 
industry.  When  the  price  has  fallen  sufficiently,  all  that  is  pro- 
duced will  be  usefully  used  and  the  substance,  gold  or  whatever 
it  may  be,  will  have  fallen  to  its  true  comparative  value,  or 
industrial  worth. 

It  must  be  borne  in  mind  that  the  prices  of  iron,  copper  and 
nickel,  as  well  as  silver  and  gold,  went  through  the  same  stages, 
from  high  artificial  price  to  real  industrial  worth. 

46.  In  the  case  of  public  utilities,  depreciation  is  not  a  sum 
of  money  paid  outright  to  the  companies,  but  is  in  reality  a 
trust  fund  to  insure  maintaining  the  service  at  par.    It  is  the 
duty  of  the  commissions  to  see  that  it  is  all  properly  used. 
But  in  taking  depreciation  into  account,  we  must  also  take 
into  account  appreciation.    To  charge  the  public  for  that  which 
depreciates  without   allowing  for  that  which  appreciates  in 
value  is  eminently  unjust.     In  private  companies  the  matter 
should  be  treated  likewise  if  the  true  status  of  the  business  is 
desired. 

47.  While  each  item  of  a  plant  has  usually  a  different  depre- 
ciation rate,  yet  a  mean  rate  for  the  entire  plant  may  be  ob- 
tained so  that  the  plant  may  be  treated  as  a  whole.    This  is 
illustrated  in  the  following  illustrative  problem. 

Example  14.  A  power  plant  of  1000  h.p.  consists  of  the  fol- 
lowing items,  with  costs  and  depreciation  rates  as  set  forth. 
Determine  the  mean  depreciation  rate. 

Solution:  Summing  up  the  costs  gives  us  a  total  of  $91,400 
for  the  plant.  So  also  summing  up  the  depreciation  amounts 
gives  us  $1682.10  —  $1280  =  $402.10  net  depreciation.  The 


42         FUNDAMENTAL  FINANCIAL  CALCULATIONS 

mean  depreciation  rate  is  evidently  the  net  depreciation  amount 
divided  by  the  total  cost,  or 

402.10  -T-  $91,400  =  0.44%  Ans. 

DEPRECIATION 


Item 

Cost 

Rate 

Amount 

Grounds. 

$16,000  oo 

—  8  o% 

—  $1280  oo 

Buildings,  concrete  

9,000.00 

O    3 

27    OO 

Turbo-generators 

1  8  ooo  oo 

2    7 

486  oo 

Condensers  

3  800  oo 

40 

1^2    OO 

Boilers  

18,000  oo 

I    O 

180  oo 

Stokers 

2   COO   OO 

6  o 

Stacks  brick.  . 

3  OOO   OO 

O    T.T, 

900 

Coal  bins,  steel  
Coal  conveyers  

1,800.00 
4,000  .  oo 

5-9 
6.0 

106.  20 
240  oo 

Boiler  feed  and  service  pumps.  .  .  . 
Feed  water  heaters,  open  
Switchboard  and  wiring  
Exciters        

800  .  oo 
1,600.00 
2,400.00 
1,700  oo 

3-0 
i  .0 
0-35 

•2     O 

24.00 
16.00 
8.40 

r  T    on 

Foundations  (mach.).  . 

1,000.00 

o 

OO   OO 

Piping  and  conduits  

5,600.00 

3  .0 

168  oo 

Crane  . 

I  2OO   OO 

O    3 

3    00 

48.   Present  Worth  of  a  Depreciating  Equipment.  —  As  an 

equipment  depreciates,  the  sum  of  its  present  worth  plus  the 
depreciation  fund  together  with  accrued  interest  thereon, 
equals  the  first  cost  of  the  equipment.  It  is  often  desirable  to 
know  the  worth  of  a  depreciating  unit  at  some  given  time. 
This  we  can  determine  as  follows: 

As  before  let 

C  =  the  first  cost, 
D  =  the  depreciation  rate, 
R  =  the  interest  rate, 
n  =  the  total  life  of  the  unit, 
and  m  =  the  actual  age  of  the  unit. 

At  the  end  of  each  year  we  lay  aside  DC  dollars.  So  at  the 
end  of  the  first  year  the  depreciation  fund  (F)  amounts  to 

Fl  =  DC. 
This  increases  to  DC(i  +  R)  at  the  end  of  the  second  year,  to 


WORTH   OF   A    DEPRECIATING   EQUIPMENT        43 

which  we  add    another  DC  dollars.     The  depreciation  fund 
therefore  amounts  to 

F2  =  DC(i  +  R)  +  DC  =  J9C[(i  +  R)  +  i]. 
So  also  at  the  end  of  the  third  year  the  depreciation  fund  is 

Fs  =  DC[_(i  +  R)2  +  (i  +  R)  +  i]. 
And  at  the  end  of  (m)  years,  it  is 


But 

(i  +  *)-i  +  (i  +  U)-2  -   .   .  +  i  = 
so  that 


But  since 


(i  +  R)n  -  i 
we  get  by  substitution 

Fm  = 


The  present  worth  (W)  is  the  first  cost,  (C),  less  the  total  depre- 

ciation (Fm)  ,  so  that 

W  =  C  -  Fm 

which,  by  substitution,  becomes 

C[(l    +*)"-!]  (Q] 


Example  15.  The  original  cost  of  a  unit  is  $10,000  and  its 
life  is  30  years.  The  interest  rate  is  5%.  What  is  the  worth 
of  the  unit  at  the  end  of  the  2oth  year? 

Solution:  In  this  case 

n  =  3°  years, 
m  =  20  years, 

R  =  5%, 
and  C  =  $10,000, 

so  that 

_  io,ooo[(i.Q5)»  -  (1.05)*] 

(i.os)30  -  i  . 
or  W  =  $5023.90. 


44         FUNDAMENTAL  FINANCIAL  CALCULATIONS 

Example  16.  A  building  has  a  life  of  50  years.  Interest  is 
4%.  What  per  cent  of  the  first  cost  is  its  worth  at  an  age  of 
40  years? 

Solution:  In  this  case 

n  =  5°  years, 
m  —  40  years, 

R  =  4%, 
C  =  100%, 
whence 


9.  Vestance.  —  The  first  cost  of  the  same  type  of  equip- 
ments varies  greatly.  The  life  of  the  various  types  and  their 
cost  of  operation  offer  two  more  intensely  variable  conditions. 
Besides  all  these  we  must  consider  the  different  classes  of  equip- 
ment that  are  all  designed  to  serve  the  same  purpose.  Thus, 
for  example,  we  have  to  consider  the  various  classes  of  pumps, 
such  as  reciprocating,  centrifugal,  and  jet  pumps.  And,  for 
any  one  class,  i.e.,  reciprocating  pumps,  we  must  consider  the 
various  types,  such  as  simplex,  duplex,  or  triplex  pumps, 
single  or  double  acting,  vertical  or  horizontal  and  so  forth. 

With  such  a  kaleidoscopic  mass  of  variable  factors,  it  would 
seem  at  first  hardly  possible  to  get  a  basis  from  which  compara- 
tive values  of  such  different  classes  and  types  may  be  deter- 
mined. Yet  such  a  basis  may  be  rather  readily  obtained  in  what 
we  call  vestance,  the  equivalent  cost  of  a  permanent  service. 

We  shall  lead  up  to  this  with  the  following  example. 

Example  17.  If  we  make  a  certain  crossing  with  a  wooden 
bridge,  it  will  cost  $10,000  and  its  life  will  be  ten  years.  If  we 
make  this  crossing  with  a  steel  bridge,  its  cost  will  be  $18,000 
and  have  a  life  of  thirty  years.  If  the  interest  rate  is  6  %,  which 
of  the  two  bridges  will  be  the  most  economical  investment? 

Solution:  The  first  cost  of  the  wooden  bridge  is  $10,000. 
At  the  end  of  ten  years  we  mus  t  again  build  the  wooden  bridge 
at  the  cost  of  another  $10,000.  Finally  at  the  end  of  twenty 


VESTANCE  45 

years,  we  must  again  rebuild  the  wooden  bridge  at  the  cost  of 
still  another  $10,000,  when,  neglecting  other  considerations, 
we  will  have  attained  the  life  of  the  steel  bridge,  30  years,  and 
rendered  with  the  wooden  bridge,  or  rather  with  three  wooden 
bridges,  the  same  service  as  with  the  steel  bridge. 

To  begin  with,  then,  we  invest  $10,000.    At  the  end  of  ten 
years  we  again  invest  another  $io,ooo,whose  present  worth  (P)  is 

10.000 


So  at  the  end  of    twenty   years  we  must   invest  another 
$10,000,  whose  present  worth  is 

n          10,000 

• 


The  total  cost  of  this  service  for  thirty  years,  with  the  wooden 
bridges  is 

10,000  +  5585  +3119  =  $18,704. 

The  same  service  may  be  rendered  with  a  steel  bridge  for 
$18,000,  showing  that  the  latter  is  the  best  by  $704.  Of  course 
to  these  costs  must  be  added  that  of  maintenance,  loss  to  traffic 
during  reconstruction,  and  such  other  considerations  as  the 
physical  nature  of  the  specific  problem  may  demand.  Taking 
these  into  consideration,  in  the  above  case,  would  throw  the 
decision  very  decidedly  in  favor  of  the  steel  bridge. 

50.  Depreciation  Vestance.  —  The  depreciation  vestance  is 
the  total  of  the  present  worths  of  the  investment  and  reinvest- 
ments. We  can  determine  this  as  follows: 

Let  C  =  the  first  cost, 

n  =  the  life  in  years, 
and  R  =  the  interest  rate. 

To  begin  with,  we  invest  (C)  dollars.  At  the  end  of  (n) 
years,  we  must  invest  another  (C)  dollars,  whose  present  worth  is 

p       c 

"     (i  +  R)" 


46         FUNDAMENTAL   FINANCIAL  CALCULATIONS 

Again  at  the  end  of  (2w)  years,  we  must  invest  another  (C) 
dollars,  whose  present  worth  is 

P2n  =  _    c 

and  so  again  at  the  end  of  (3^)  years,  we  must  invest  another 
(C)  dollars,  whose  present  worth  is 

p  -    c 

n    ~~( I       P\3n 

and  so  on.  So  in  general  at  the  end  of  (mn)  years  we  must 
again,  for  the  wth  time,  invest  (C)  dollars,  with  a  present 
worth  of 

p  -     c 

mn  —  7 — T" — pWn' 
Summing  all  these  up  gives 
,  v     r  ,        C  C  C_  r 

Vd=         t"   (T_i-^n   +    (._±_»\2n   +    fT_|_ltt»n    '      '      '  + 

or 


Let  (S)  equal  the  sum  of  the  series  in  the  brackets.    Then 

c  _  T    .          I          .          I  ,          i 

*•*  I        /  T-»\  „       I        /  i         -r>\  o_  I        /  T->\  .. 


Reducing  to  a  common  denominator,  gives 


«      V-1  •*•»•/  i^V-1-    i^-1*-/  I^V-1!^ •*•*•/       T^V-1    I  •*»-/  •       /     v 

Now  multiply  through  by  (i  +  R)n  and  we  get 

(i+R)nS= —  — (i  +I#)"m  ~~'      ^ 

And  subtracting  equation  (a)  from  equation  (b)  gives 

(,  D\nC          O     __     (J     +    1?)(*+1>»-     I 


and 

"  (i  +  £)*»[( i  +  /?)»  -  i]' 


DEPRECIATION  VESTANCE  47 

Now  dividing  through  by  (i  +  R)mn  gives 


s  = 


(i  +  £)n  -  i  (i  +  #)w  -  i 

When  (inn),  the  length  of  the  service  becomes  infinite,  then 
(i  +  R)mn  also  becomes  infinite  and  -.  —  becomes  zero. 

So  for  the  complete  summation  (m  =  oo)  of  the  above  series, 
we  get 

(i  +  *)• 
"  (i  +  R)n  -  i' 
And  since 

vd  =  cs 

we  have 


But  we  had  that  ¥-t  —  '    ~  T  is  the  term  factor  (T),  so  that 

Fd=        ..........  (n) 


Example  18.  A  concrete  building  has  a  first  cost  of  $100,000 
and  a  life  of  50  years.  If  the  interest  rate  is  5%,  what  is  the 
depreciation  vestance? 

Solution:  In  this  case 

C  =  $100,000, 

R  =  5%, 

and  n  =  50  years, 

so  that  by  Table  2 

T  =  0.91279, 

and  the  depreciation  vestance  is 

100,000       fa 

Vd  =  -  -  -  =  $109,550  Ans. 
0.91279 

If  the  interest  rate  were  only  3  %,  then  the  term  factor  would 
be,  by  Table  2,  only  0.77189  and  the  depreciation  vestance 
would  be 

rr  IOO.OOO         „,, 

Vd  =  -      -—  =  $129,540. 
0.77189 


48        FUNDAMENTAL  FINANCIAL  CALCULATIONS 

Example  19.  A  wooden  building  to  take  the  place  of  the 
concrete  building  of  problem  (18),  would  cost  $70,000  and  have 
a  life  of  20  years.  With  interest  at  5  %,  which  would  be  the  best 
investment? 

Solution:  Here 

C  =  $70,000, 
n  =  20  years, 

R  =  5%, 
so  that 

T  =  0.62312, 
and 

T7         70.000        a. 
Vd  =  -t-z    —  =  $112,340. 
0.62312 

This  would  make  a  difference  of 

112,340  -  109,550  =  $2790 

in  favor  of  the  concrete  building,  even  excluding  the  higher 
maintenance  cost  of  the  cheaper  structure. 

If  the  interest  rate  were  3%,  then  the  term  factor  (T) 
would  be  0.44632  and  the  depreciation  vestance  would  be 


Vd  =  .  =  $I56,838. 

0.44632 

In  this  case  the  difference  would  be 

156,838  -  129,540  =  $27,298. 

Example  20.     A   permanent    concrete   viaduct   costs  $5000. 
What  is  its  depreciation  vestance? 

Solution:  In  this  case  n  =  oo  so  that  the  term  factor  is 
unity.    The  depreciation  vestance  is  then 

=  $5000, 


the  same  as  the  first  cost. 

61.  Operating  Vestance.  —  With  depreciation  vestance,  we 
get  a  method  of  comparing  equipments  having  different  first 
costs  and  different  lives.  We  thus  have  a  method  of  determin- 
ing comparative  values  as  to  fixed  charges  and  those  only. 


OPERATING  VESTANCE  49 

The  operating  costs  are  continuous,  and  are  not  limited  to  a 
term  of  years.    By  capitalizing  the  annual  cost  of  operation, 
we  get,  therefore,  the  operating  vestance. 
Letting 

a  =  operating  cost  per  year,  and 
R  =  interest  rate, 
then 

V.  =  I  .........  (12) 

This  is  on  the  assumption  that  the  operating  cost  is  constant. 
However,  improvement  in  design  is  what  renders  equipments 
obsolete,  and  this  refers  particularly  to  reductions  in  the  cost  of 
operating.  This  more  than  compensates  for  obsolescence  and 
both  it  and  reduction  in  cost  of  operation  must  be  taken  into 
consideration. 

52.  Total  Vestance.  —  The  total  vestance  (F)  is  evidently 
the  sum  of  the  depreciation  and  operating  vestance,  so  we  have 

V   =    Vd  +    Va 


or 


or 


So  that  knowing  the  life,  operating  cost  and  vestance,  we  can 
determine  what  the  cost  should  be.     Also  since 


(I3B) 


we  can  determine  the  life  of  the  unit,  knowing  the  first  cost, 
vestance  and  operating  cost.  In  a  similar  manner  we  can  solve 
for  any  of  the  other  quantities  in  equation  (13). 

Example  21.  A  1000  h.p.  Diesel  Engine  plant  costs  $75,000. 
The  life  of  the  engine  is  20  years.  The  interest  rate  is  5%. 
The  cost  of  operation  is  0.4  cent  per  horse  power  hour  (h.p.h.). 


50        FUNDAMENTAL  FINANCIAL  CALCULATIONS 

If  the  engine  is  operated  at  full  load  for  3000  hours  per  year, 
what  will  be  the  vestances? 

Solution:  Here 

C  =  $75,000, 

R  =  5%, 
n  =  20  years, 
so  that 


and 

Vd  = 


0.62312 


The  total  annual  operating  cost  is 

1000  X  0.004  X  3000  =  $12,000, 
so  that  the  operating  vestance  is 

T7        12,000       .., 

Va    =  -  -  --    =    $240,000. 
0.05 

The  total  vestance  is  therefore 

120,360  +  240.000  =  $360,360. 

Example  22.  If,  in  Example  20,  we  put  in  a  steam  plant,  the 
first  cost  will  be  $50,000,  and  the  life  is  likewise  20  years,  while 
the  cost  of  operation  will  be  0.7  cent  per  h.p.h.  With  the 
same  interest  rate,  i.e.,  5%,  which  will  be  the  best  investment? 

Solution:  In  this  case 


Vd  .  --  =  $80,240. 


The  annual  cost  of  operation  is 

1000  X  .007  X  3000  =  $21,000. 
The  operating  vestance  therefore  is 
,,        21,000       4. 

Va    =  --    =    $420,000, 
0.05 

giving  a  total  vestance  of 

420,000  -f-  80,240  =  $500,240. 


TOTAL  VESTANCE  51 

The  difference  between  the  total  vestance  of  the  steam  and 
Diesel  plant  is 

500,240  -  360,360  =  $139,880 

in  favor  of  the  Diesel  plant. 

Example  23.  If  the  power  were  purchased  at  one  cent  per 
h.p.h.  and  the  motor  cost  $6000  and  had  90%  efficiency  with  a 
life  of  30  years,  how  would  the  total  vestance  compare  with 
that  of  the  steam  and  Diesel  plants? 

Solution:  In  this  case 

n  =  30  years, 
C  =  $6000, 

R  =  5%, 
so  that 

T  =  0.76861, 


and  Vd  =  --   =  $7810. 

0.76861 

The  total  annual  cost  of  operation  is 

loop  X  o.oi  X  3000  _  <n;7 
-  *3 
0.90 

and  the  operating  vestance  is 

Va  =  33.333  =  $666,667 
0.05 

giving  a  total  vestance  of 

7810  +  666,667  =  $674,477. 
This  gives  a  difference  of 

674,477  -  500,240  =  $174,237 
in  favor  of  the  steam  plant,  and  a  difference  of 

674,477  -  360,360=  $314,117 
in  favor  of  the  Diesel  plant. 

Example  24.  At  what  price  would  the  electric  power  in 
Example  23  have  to  be  purchased  in  order  that  the  electric 
motor  plant  may  equal  (a)  the  steam  plant  of  Example  22 
and  (b)  the  Diesel  plant  of  Example  21? 


52        FUNDAMENTAL  FINANCIAL  CALCULATIONS 

Solution:  Let  p  =  the  price  of  the  electric  power  per  h.p.h. 
Then  the  annual  cost  of  operation  is 

1000  X  p  X  3000 

Q  -  =  3,333,333^; 

the  operating  vestance  is 

Va  =  3,333,3331  =  $66,666,65y^ 
0.05 

and  the  total  vestance  is 

V=  $66,666.667^  +  $7810. 

(a)  If  the  total  vestance  of  the  motor  and  steam  plant  are 
equal,  then 

66,666,667/7  +  7810  =  $500,240, 
so  that 

p  =  $0.0074 

or  a  little  over  0.7  of  a  cent. 

(b)  If  the  total  vestance  of  the  motor  and  Diesel  plants  are 
made  equal  then 

66,666,667/7+  7810=  $360,360 
so  that 

p  =  $0.0053 

or  a  little  over  \  cent  per  h.p.h. 
53.   Annual  Operating  Cost.  —  If  we  call 

L  =  the  load  in  h.p., 

a  =  the  operating  cost  per  h.p.h. 
and 

N  =  the  hours  of  operation  per  year, 

then,  if  (L)  and  (a)  are  constant,  the  total  operating  cost  per 
year  (^4)  is 

A  =  LaN       .....'  .....    (14) 

But  if  the  load  and  operating  cost  vary,  the  annual  cost  of 
operation  will  be 

86o 

LadN   .    ...    .    .     (15) 


/ 


the  upper  limit  (8760)  being  the  total  number  of  hours  in  the 


ANNUAL  OPERATING  COST  53 

year  and  therefore  the  maximum  number  of  hours    per  year 
that  any  unit  may  operate. 

Example  25.  For  a  certain  power  plant,  we  find  that  the  load 
variation  with  the  hours  of  operation  may  be  expressed  by  the 
following  approximate  equation 

L  —  7000+  o.2N  —  'o.oooiA^2, 

while  the   operating  cost  per  h.p.h.  varies  with  the  load  in 
accordance  with  the  following  approximate  equation 

a  =  0.008  —  0.0000004!,. 
What  will  be  the  total  annual  operating  cost? 

Solution:  Substituting  in  the  equation  for  (a),  the  value  of 
(L)  in  terms  of  (N)  we  get 

a  =  0.008  —  0.0000004  (7000  -f-  o.2N  —  o.oooiN2) 
or 

a  =  0.0052  —  0.00000008^  +  0.00000000004  A7"2 
whence 

La  =  (7000  +  o.2N  —  o.oooiA7'2)  (0.0052—  o.oooooooS^V 

+  0.00000000004  N2) , 
or 

La  =  (36.40  +  o.oo^SN  —  0.00000024^  +  o.ooooooooooi6]V3 
—  0.000000000000004  A^4) . 

Substituting  this  in  equation  (15),  we  get 

$760 
A  =    I        LadN 


f 

Jo 


so  that  in  this  case 

A  = 

[36.40^  +  o.ooo24Ar2  —  0.00000008053$*. 

+  0.000000000004  N4  +  o.ooooooooooooooo 
or 

A  =  318,864  +  18,417.26  -  54,134.69  +  23,544.94  -  41,268.26, 

so  that 

A  =  $265,433.25  Ans. 

54.  In  any  actual  problem  in  practice,  the  load  assumed  is 
that  which  is  actually  carried,  according  to  the  records,  or  that 
which,  in  the  light  of  previous  experience,  may  be  anticipated. 


54         FUNDAMENTAL  FINANCIAL  CALCULATIONS 

The  curve  for  the  variation  of  the  operating  cost  with  the 
load  is  obtained  either  from  previous  experience  or  from  the 
factory  guarantee. 

In  the  examples  given,  it  appears  that  the  operating  vestance 
far  exceeds  the  depreciation  vestance.  Particularly  is  this  true 
of  equipments  with  low  first  cost.  We  can  materially  reduce 
the  total  vestance  in  many  systems  and  therefore  the  produc- 
tion cost  by  putting  in  equipment  with  higher  efficiency  and 
allowing  a  greater  first  cost  than  we  are  accustomed  to.  How- 
ever, as  we  shall  see,  conditions  are  sometimes  just  the 
opposite,  i.e.,  in  cases  we  sometimes  allow  far  too  much  for  first 
cost. 

So  far  in  vestance  we  have  taken  into  consideration  only 
interest  and  depreciation.  There  are,  however,  the  two  other 
factors  of  fixed  charges,  namely  taxes  and  insurance.  These 
factors  run  continuously,  i.e.,  they  are  not  limited  to  the  life 
of  the  equipment.  Their  vestance  may,  therefore,  be  found  by 
simply  capitalizing  the  amount  of  tax  and  insurance. 

Letting    /  =  the  amount  of  insurance  per  year, 
X=  the  amount  of  taxes  per  year, 
R  =  the  interest  rate, 

Vi  =  the  insurance  vestance, 
Vx  =  the  tax  vestance, 
then 

X 


and  Vi  =  -g 

The  total  vestance  is  then 

V  =  Vd  +  Vx+  V<  +  Va 

C   ,  X  ,   I       A      C    ,   (X+I.+  A) 


or 


V  -^  +      + 
'^'^' 


-      - 

R      T  R 


Example  26.     Determine  the    total  vestance  of  a  wooden 
building  under  the  following  conditions:    First  cost  $10,000, 


ANNUAL  OPERATING   COST  55 

life  20  years,  interest  rate    7%,    taxes   0.8%,    insurance  i% 
and  maintenance  and  repairs  $120  per  year. 

Solution:  The  term  factor  under  these  conditions  is  0.74158 
so  that 

,r  IO,OOO  rfK 

;        ..    •         v<  '  0-^8  -  $I3'200' 

X  =  10,000  X  0.008  =    80, 

/  =  10,000  X  o.oi     =  100, 

and  A  =  120 

Total  =$300 

VIAX  =  ^  =  $4286. 
0.07 

The  total  vestance  is  then 

V  =  13,200  +  4286  =  $17,486. 

Taxes  affect  all  items  of  equal  value  alike,  but  the  insurance 
varies  with  the  hazard.  An  isolated  concrete  structure, 
having  within  it  only  incombustible  material,  need  carry  no 
insurance.  But  a  wooden  building  in  a  district  with  much  fire 
hazard  must  bear  a  heavy  insurance  rate,  due  both  to  itself 
and  its  surroundings,  reducing  greatly  its  comparative  value. 

In  a  power  plant,  we  almost  invariably  have  fireproof  equip- 
ment, such  as  engines,  boilers,  feed- water  heaters,  pumps,  etc., 
housed  in  a  building  which  may  not  itself  be  fireproof.  If  the 
structure  burns  down,  this  equipment  would  be  destroyed. 
It  is,  therefore,  necessary  to  carry  insurance  on  this  machinery. 
The  question  arises  as  to  whether  the  cost  of  insurance  of 
this  equipment  should  be  charged  against  the  equipment  or 
against  the  building.  Again  we  may  have  patterns  stored  in 
a  building  which  is  itself  quite  fireproof.  The  patterns  are 
highly  combustible.  If  they  burn  they  usually  destroy  the 
building. 

In  any  such  case  it  is  evident  that  the  combustible  items 
should  be  charged  with  the  insurance  on  all  that  might  be 
destroyed  with  the  burning  of  these  items.  In  the  case  of  the 
power  plant,  the  insurance  on  the  machinery  housed  should  be 
charged  against  the  building,  while  in  the  second  instance 


56         FUNDAMENTAL  FINANCIAL  CALCULATIONS 

cited  above,  the  insurance  on  the  building  should  be  charged 
against  the  patterns  stored.  But  that  part  of  the  insurance 
which  is  collected  because  of  conflagration  hazard  should  be 
charged  against  the  location,  i.e.,  the  lot  itself,  because  this 
cost  is  entirely  due  to  its  location. 

This  much  is  definite.  But  how  should  the  insurance  cost  be 
apportioned  where  both  the  building  and  the  things  stored  are 
combustible?  This  does  not  lead  to  a  definite  solution.  The 
charge  should  be  apportioned  to  the  hazard  cause  by  each, 
i.e.,  the  structure  and  the  items  housed  and  this,  for  want  of 
sufficient  data,  remains  a  matter  of  judgment  based  on  such 
experience  as  we  now  have. 

PROBLEMS 

1.  How  much  will  $i  amount  to  in  500  years  at  4%  compound  interest? 

2.  What  is  the  present  worth  of  $5000  due  in  20  years? 

3.  In  how  many  years  will  $1000  quadruple  itself  at  8%  interest? 

4.  A  building  is  sold  on  the  following  terms:    The  selling  price  of  the 
building  is  $4000  and  is  to  bear  interest  at  the  rate  of  7%  on  all  unpaid 
amounts.     The  purchaser  is  to  pay  at  the  rate  of  $50  per  month.     From 
this  amount  interest  is  paid  and  all  above  interest  is  to  apply  towards  a 
reduction  of  capital  due. 

(a)  In  how  many  years  will  the  purchaser  complete  payment?  (b)  How 
much  money  will  the  purchaser  have  paid  the  seller  when  payment  is  com- 
pleted? (c)  If  the  building  depreciates  at  the  rate  of  7%  per  year,  how 
much  is  the  building  worth  when  payment  is  completed? 

5.  A  ten  h.p.  induction  motor  of  87%  efficiency  cost  $157.    How  much 
could  we  afford  to  pay  for  a  motor  of  88.5%  efficiency  if  the  power  costs 
2  cents  per  h.p.h.  and  the  motor  is  to  be  run  at  full  load  for  3000  hours 
per  year?    Assume  the  life  of  the  motor  as  25  years  and  interest  at  7%. 

6.  A  1000  h.p.  Diesel  engine  is  guaranteed  to  operate  on  0.40$  of  fuel 
oil  per  h.p.h.     The  engine  operates  at  full  load  4000  hours  per  year.     The 
cost  of  the  engine  is  $52,000.    Interest  is  5%.    Fuel  costs  $1.60  per  barrel 
of  320$  and  the  life  of  the  engine  is  20  years,  while  it  actually  uses  0.43$ 
of  fuel  oil  per  h.p.h.    How  much  should  be  deducted  from  the  purchase 
price  to  compensate  the  purchaser  for  this  loss  in  economy? 

7.  A  150  h.p.  220  volt  motor  costs  $1170  and  is  guaranteed  at  91% 
efficiency.    The  motor  has  a  life  of  25  years  and  is  to  be  used  for  2500 
hours  per  year.    Interest  rate  is  6%  and  power  costs  i  cent  per  h.p.h. 
If  instead  we  purchased  at  the  same  price  and  other  conditions  a  440  volt 
motor  with  only  90%  efficiency,  how  much  would  we  lose  thereby? 


PROBLEMS  57 

8.  A  storage  battery  costs  $75  per  kw.h.  of  capacity.    Its  life  is  assumed 
at  3  years,  its  efficiency  at  65%,  other  operating  costs  zero.    Interest  rate 
6%.    What  is  the  total  vestance  of  the  storage  battery  per  kw.h.,  if  it  is 
used  at  full  load  for  2000  hours  per  year,  and  electric  power  costs  i  cent  per 
kw.h.? 

9.  A  permanent  concrete  viaduct  costs  $60,000.     With  interest  at  7%, 
determine  the  total  vestance  of  this  structure.    What  would  be  the  vestance 
if  the  interest  rate  was  4%? 

10.  Calculate  the  depreciation  rate  of  an  engine  whose  life  is  21  years 
with  interest  at  6.5%. 

11.  Calculate  the  depreciation  rate  of  a  stack  whose  life  is  42  years  with 
interest  rate  at  4.5%. 

12.  An  electric  motor  costs  $300  and  has  a  life  of  30  years.    Determine 
its  worth  at  the  end  of  o,  5,  10,  15,  20,  25,  and  30  years.    Assume  an  interest 
rate  of  6%.    Plot  the  curve. 

13.  A  Diesel  engine  cost  $75,000  and  has  a  life  of  15  years.     Determine 
its  worth  at  the  end  of  o,  3,  6,  9,  12,  and  15  years.    Assume  5%  interest 
rate.    Plot  the  curve. 

14.  Which  would  be  the  best  investment,  a  wooden  building  whose  life 
is  ten  years  or  a  steel-brick  building,  costing  twice  as  much,  and  having  a 
life  of  40  years?    Assume  interest  5%,  and  operating  costs  zero. 

15.  In  problem  (14)  what  would  have  to  be  the  life  of  the  wooden  build- 
ing in  order  to  equal  the  steel-brick  building  in  worth? 

16.  A  steam  plant  has  a  first  cost  of  $70,000  and  a  guaranteed  life  of 
25  years.     The  cost  of  attendance  is  0.4  cent  per  h.p.h.,  maintenance 
o.i  cent,  and  repair  o.i  cent  per  h.p.h.     With  interest  rate  6%,  what  will 
be  the  vestances  of  this  plant  if  it  is  used  for  4000  hours  per  year  at  full 
load? 

17.  To  operate  a  certain  pump  at  full  load  for  1500  hours  per  year  we 
can  use  a  10  h.p.  gas  engine  costing  $500  and  having  a  life  of  20  years. 
The  operating  costs  for  this  engine  are  one  cent  per  h.p.h.     An  electric 
motor  to  take  the  place  of  this  engine  would  cost  only  $150  and  have  a 
guaranteed  life  of  30  years,  and  an  efficiency  of  88%.    The  power  company 
offers  power  at  2  cents  per  h.p.h.    Would  the  gas  engine  or  motor  be  the 
best  investment  under  these  conditions,  assuming  other  operating  costs  as 
zero?    What  would  be  the  difference  in  worth? 

18.  If  the  electric  power  is  offered  at  one  cent  per  h.p.h.  in  problem  (17), 
then  which  would  be  the  best  investment?     What  then  would  be  their 
differences  in  worth? 

19.  In  problem  (17),  how  much  could  you  pay  the  power  company  so 
that  the  comparative  value   (vestance)   of  the  two  propositions  will  be 
equal? 

20.  In  problem  (17),  if  the  pump  was  to  be  operated  only  1000  hours 
per  year  which  would  be  the  best  investment? 


58        FUNDAMENTAL  FINANCIAL  CALCULATIONS 

21.  In  problem  (17),  if  the  pump  was  to  be  operated  only  500  hours  per 
year,  which  of  the  two  plants  would  be  the  best  investment? 

22.  In  problem  (17),  for  how  many  hours  of  operation  would  the  com- 
parative value  of  the  two  plants  be  equal? 

23.  For  a  certain  plant,  the  load  in  h.p.  varies  according  to  the  follow- 
ing equation: 

L  =  12,000  —  o.^Nj 

and  the  operating  cost  per  h.p.h.  varies  according  to  the  equation 
a  =  0.007  —  O.OO00005JV. 

What  will  be  the  total  annual  cost  of  operation? 

24.  For  a  certain  plant  the  load  in  h.p.  varies  according  to  the  equation 

L  =  500+  o.2N. 

and  the  operating  cost  per  h.p.h.  varies  according  to  the  equation 
a  =  0.004  ~  0.00000027V. 

What  will  be  the  total  annual  cost  of  operation? 

25.  A  structure  costs  $60,000  and  has  15  years  life.     Interest  rate  is 
4%,  insurance  i%,  taxes  0.7%,  cost  of  maintenance  and  repair  1.2%. 
Determine  the  total  vestaiice  of  this  structure. 

26.  In  problem  (25)  how  much  could  we  afford  to  pay  for  a  permanent 
fireproof  structure  whose  cost  of  maintenance  and  repair  was  0.5%? 


CHAPTER  IV 
BASIC  COSTS 

First  and  Operating  Costs  at  Full  and  Fractional  Loads  of  Steam  Engines, 
Boilers,  Buildings,  Centrifugal  Pumps,  Induction  Motors,  Direct 
Current  Motors,  Generators,  Transformers,  Oil  Engines,  Producer 
Gas  Engines,  Diesel  Engines,  Producers,  Standard  Pipe,  Casing,  Wood 
and  Riveted  Steel  Pipe.  Cost  of  Tunnels,  Canals,  Excavations, 
Hydro-electric  Installations,  Compressors,  Condensers,  Economizers, 
Fans,  Feed  Water  Heaters,  Stokers,  Water  Treating  Plants,  Super- 
heaters and  Turbines. 

55.  Basic  Costs.  —  Every  science,  in  its  practical  applica- 
tion, is  based  upon  experimental  data.  These  data  are  not 
absolutely  exact  but  usually  are  exact  enough  to  give  good 
working  results.  In  electricity,  for  example,  we  use  in  our  cal- 
culations the  experimental  values  of  such  constants  as  the  co- 
efficients of  resistance,  inductance,  and  the  like.  The  theory 
is  absolutely  correct.  But  these  constants  are  only  closely 
approximate.  In  heat  we  have  such  constants  as  the  specific 
heat  of  the  various  substances,  latent  heats,  fuel  values,  and 
the  like,  determined  by  experiment.  So  in  financial  engineer- 
ing we  have  for  our  "  experimental "  data,  the  basic  costs  of  our 
equipment,  structures,  and  the  like,  of  materials  consumed, 
such  as  fuel,  oil,  iron,  and  the  like  and  finally  of  our  costs  of 
attendance  and  maintenance.  We  must  know  how  these  costs 
vary  with  differences  in  size,  class,  and  type  of  the  equipments, 
and  the  like.  These  data  will  only  be  closely  approximate. 
But  as  their  importance  in  engineering  is  better  understood, 
more  effort  will  be  expended  in  getting  still  better  data.  Cost 
data  are,  however,  more  uniform  and  accurate  than  other 
scientific  data. 

It  seems  peculiar  and  is  indeed  very  unfortunate  that  so 
many  authors  in  their  engineering  books  give  no,  or  very  little, 

59 


60  BASIC  COSTS 

consideration  to  costs,  in  spite  of  the  fact  that  the  primary 
duty  of  the  engineer  is  to  consider  costs  in  order  to  attain  real 
economy  —  to  get  the  most  power,  for  example,  not  from  the 
least  number  of  pounds  of  steam,  but  from  the  least  possible 
number  of  dollars  and  cents:  to  get  the  best  financial  efficiency. 
Two  very  notable  exceptions  to  the  above  are  Fernold  and 
Orrok's  "  Engineering  of  Power  Plants  "  in  which  a  very  great 
deal  of  valuable  engineering  data  is  given,  and  Harding  and 
Willard's  "  Mechanical  Equipment  of  Buildings."1 

56.  Records.  —  The  object  of  keeping  records  is  to  furnish 
data  for  the  engineering  staff  from  which  they  may  devise 
means  of  improving  the  economy  of  the  system,  and  of  predict- 
ing the  costs  of  future  operations.  But,  in  many  organizations, 
the  recording  or  so-called  accounting  department  is  made 
independent  of,  instead  of  subsidiary,  to  the  engineering  divi- 
sion. Records  are  of  absolutely  no  value  unless  they  are  studied 
and  conclusions  drawn  therefrom  that  will  aid  in  getting  still 
further  improved  economy.  This  is  well  stated  by  Gebhardt l 
in  " Steam  Power  Plant  Engineering"  when  he  says:  "Few 
engineers  realize  the  importance  of  a  detailed  system  of  account- 
ing, or  the  saving  which  may  be  effected  in  cost  of  operation  by 
careful  study  of  the  daily  records  of  performance.  Many 
regard  graphical  load  curves,  meter  readings,  and  similar  records 
as  interesting,  but  of  little  economic  value.  During  the  past 
few  years,  the  author  has  made  a  close  study  of  the  cost  of 
power  in  a  large  number  of  central  and  isolated  stations  in 
Chicago,  and  found,  without  exception,  that  the  highest 
economy  was  effected  by  the  engineers  who  kept  the  most 
systematic  records;  the  poorest  results  were  obtained  where 
records  were  kept  indifferently  or  not  at  all.  In  some  small 
plants  the  numerous  duties  of  the  engineer  prevented  him  from 
devoting  the  necessary  time,  but  in  the  majority  of  cases  the 
absence  of  records  was  due  entirely  to  a  lack  of  interest.  Power 
plant  records  to  be  of  value  must  be  closely  studied  with  a  view 
to  improvement.  The  mere  accumulation  of  data  to  be  filed 
away  and  never  again  referred  to  is  a  waste  of  time  and  money." 
1  John  Wiley  &  Sons,  Inc. 


RECORDS  61, 

Where,  however,  the  bookkeeping  is  independently  done, 
instead  of  under  the  supervision  of  the  engineers,  it  is  usually 
unavailable  both  as  to  form  and  location. 

57.  In  basic  costs,  we  have  to  consider  the  variation  of  first 
costs  of  different  equipment  with  size,  the  variation  of  attend- 
ance and  economy  in  materials  consumed  with  load,  as  well  as 
the  interrelation  of  one  part  of  a  system  with  another  as  to 
costs.  Thus,  if  we  do  not  put  in  condensers  in  a  steam  plant, 
we  will  reduce  our  first  costs  by  this  amount.  But  at  the  same 
time,  due  to  the  increased  steam  consumption  of  the  engines, 
it  will  be  necessary  to  increase  the  size  of  the  boilers,  steam 
pipe,  and  the  like,  often  by  some  30  or  40%;  the  increased  cost 
of  these  items  may  far  more  than  offset  the  cost  of  the  con- 
densers. So  also  in  a  pumping  system,  a  reduction  in  the  size 
of  the  pipe  will  make  it  necessary  to  install  a  larger  motor  or 
engine  to  drive  the  pump,  due  to  increased  friction  head, 
increasing  not  only  the  first  cost  of  this  part  of  the  system, 
but  the  operating  cost  as  well.  And  so  for  almost  any  other 
item  of  equipment.  In  thus  attempting  to  "save  "  money,  care 
must  be  used  that  the  cost  of  the  saving  does  not  exceed  the 
amount  saved. 

It  is  well  known  that  there  is  an  enormous  difference  in  the 
cost  of  apparatus  for  rendering  a  given  service.  A  jet  pump 
will  deliver  water  just  as  a  centrifugal  or  plunger  pump  will. 
And  it  usually  costs  only  2  or  3  %  ?s  much.  Yet  the  difference 
in  economy  more  than  warrants  purchasing  the  more  expensive 
types.  The  cost  of  steam  engines  varies  from  $7  per  h.p.  for  a 
simple,  high-speed  engine  to  $40  per  h.p.  for  a  horizontal 
four- valve  engine.  The  question  naturally  arises  just  how  much 
can  be  paid  for  a  given  unit.  It  is  by  no  means  true  that  the 
most  expensive  is  the  best.  The  test  is  of  course  a  comparison 
of  the  total  vestance  of  both  equipments  for  anticipated  load 
conditions. 

Off-hand  the  giving  of  prices  seems  a  rather  hopeless  under- 
taking. There  is  first  the  variation  in  price  of  the  same  article 
of  practically  equal  quality  as  between  various  manufacturers, 
due  to  their  lack  of  knowledge  of  exact  unit  costs;  the  varia- 


62  BASIC   COSTS 

tion  of  price  as  between  different  localities,  and  the  variation 
of  price  with  speed,  type,  and  class  of  equipment.  Yet,  if  in 
choosing  cost  for  our  curves,  we  confine  ourselves  in  speed, 
type,  and  so  forth,  to  standard  practice,  as  we  have  done  in  the 
following  curves,  the  matter  simplifies  itself  greatly. 

Again  there  is  a  continual  variation  in  price  with  time. 
Besides  short  fluctuations  which  may  be  either  up  or  down, 
there  is  a  continual  tendency  for  the  price  to  increase  with 
time  as  measured  by  the  weight  of  gold  necessary  to  effect  a 
purchase.  This  is  due  to  the  continual  increase  in  the  gold 
reserve,  due  to  the  fact  that  more  gold  is  annually  produced 
than  is  used  in  the  industries.  There  must  therefore  be  a 
decrease  in  the  worth  of  the  gold  and  thus  an  increase  in  price 
of  commodities.  Besides  this,  there  are  price  variations  due  to 
extraordinary  conditions  as  of  war.  However,  the  worth  of  the 
units,  or  their  comparative  values  do  not  change  except  where 
further  improvements  are  made  or  new  properties  and  uses 
are  found  and  developed.  With  the  exception  of  the  last,  a 
change  of  price  does  not  change  the  size  of  a  one  h.p.  motor, 
nor  is  the  food  value  of  a  pound  of  wheat  altered  though  the 
price  doubles. 

In  any  event,  the  following  data  are  intended  primarily  as  a 
first  approximation  and  as  illustrative  of  the  method  of  applica- 
tion of  the  technique  of  financial  engineering  to  concrete 
examples.  In  any  specific  case,  exact  data  may  be  readily 
obtained.  The  prices  given  below  may  be  converted  to  prices 
as  of  today,  by  dividing  them  by  the  comparative  value  of  the 
dollar. 

58.  Steam  Engines.  —  The  average  price  of  the  principal 
types  of  steam  engines  is  given  below. 

TABLE  4 

SIMPLE  HIGH  SPEED  ENGINES  —  ERECTED 

AVERAGE  1906-10  PRICES 

Non-condensing 

I.h.p 20          30         40         50         75         TOO        150       200       250 

Price  per  i.h.p.  .$34.75  $25.60  $21.00  $18.25  $14-60  $12.75  $10.90  $10.00  $9.40 


STEAM  ENGINES  63 

The  equation  for  these  values  is  approximately 


or  C  =  550  +  7.25M    ..........   (17) 

where  P  =  the  price  per  i.h.p., 
and     M  =  the  rated  size,  in  i.h.p., 
C  =  the  price  of  the  engine. 

TABLE  5 

SIMPLE  CORLISS  ENGINES  —  ERECTED 
1906-10  PRICES 
Non-condensing 

I.h.p 100          125  150         175         200        250        300         350 

Price  per  i.h.p ....  $21.50  $20.00  $19.00  $18.25  $I7-75  $17.00  $16.50  $16.15 

I.h.p 400          450  500 

Price  per  i.h.p ....  $15.87  $15.67  $15.50 

The  equation  for  these  values  is 

P  =  1^.  +  14 (18) 

C  =  750  +  *4M (19) 

TABLE  6 

COMPOUND  ENGINES  —  ERECTED 

1906-10  PRICES 

Condensing 

I.h.p loo          200  300        400        500        600         700 

Price  per  i.h.p $33«o°  $25.00  $22.33  $21.00  $20.25  $19-67  $iQ-43 

I.h.p 800          900  looo       1500       2000 

Price  per  i.h.p $19.00  $18.78  $18.60  $18.07  $17.80 


The  equation  for  these  values  is 
1600    , 


f         ^ 
(20) 


C  =  1600  +  ifM (21) 


64 


BASIC   COSTS 


TABLE   7 
TURBO-GENERATORS 

1906-10  PRICES 
Direct  connected  a.  c.  60  cycle,  condensing 

Kv.a 5°° 

Price  per  kv.a $21 .90 

Kv.a 

Price  per  kv.a 


750 

$18.23 

5000 

$12.00 


1000 
$16.40 

75OO 
$11.63 


2000 
$13.65 

IO,OOO 
$11.45 


3000 
$12.73 


The  equation  for  these  values  is 

C  =  55.00  +  io.goM 

p  =  55-Qo 
M 


-h 


(23) 


SM& 
30 

5520 

I 
|i, 

10 

5 

i 

\ 

V 

\ 

^Co 

npou 

DdC 

ndei 

ain* 

Eng 

nes 

\ 

\ 

\ 

"^ 

^ 

V 

^5 

• 

•  • 

•      • 

•*•»« 

5 

^ 

*^ 

=* 

-Sim 
C 

Ple^ 
orlis 

on-c< 
En{ 

^"^**^ 

nden 
ines 

*  , 
3  ruff 

*•'   •> 

*•    — 

— 

*«^^s 

=^ 

^Din 

••'  •. 

as 

••  • 

nnec 
nera 

—  — 

ed 
ors- 

—  — 

^ 

Er« 

Si 

nple 

Hist 

Er 

-spet 

pcted 

"dEi 

gine 

0          200        400         600        800        1000      1200       1400       1600       1800      20( 
I.H.P. 

FIG.  i.  — Showing  the  variation  in  cost  per  i.h.p.  of  steam  engines  and 
turbines  with  size  and  type. 

59.  Operating  Costs.  —  The  number  of  pounds  of  dry  sat- 
urated steam  used  per  indicated  horse  power  hour  (i.h.p. h.) 
for  compound  condensing  engines  with  26-inch  vacuum  and 
ioo#  gage  initial  pressure  is  as  given  in  table  8: 


OPERATING  COSTS    \  65 

TABLE  8 

STEAM  CONSUMPTION 

COMPOUND  CONDENSING  ENGINES 

I.h.p  .............     50        60        70        80        100  125       150       200       250 

Lbs.  per  i.h.p  .....    $20.2  $19.  6  $19.1  $18.7  $18.0  $17.5  $17.3  $16.7   $16.4 

I.h.p  .............       300      400      500      600       700  800      900      1000 

Lbs.peri.h.p  ......  $16.1  $15.7  $15.4  $15.2  $15.1  $15.0  $14.9  $14.8 

• 

These  values  correspond  to  the  equation 

13-2,  (24) 


where  W  =  the  pounds  of  steam  per  i.h.p.h., 
and      M  =  the  size  in  i.h.p. 

With  fractional  loads  the  steam  consumption  is  as  given  in 
the  following  tables.  In  these  tables,  only  the  strokes  of  the 
engines  are  given.  Since  the  allowable  piston  speed  is  well 
known,  the  stroke  determines  the  r.p.m.  The  i.h.p.  for  any  of 
these  engines  may  therefore  be  readily  determined,  for  any 
given  bore,  which  may  be  anything  from  one-half  to  one  times 
the  stroke.  A  stroke  equal  to  ij  times  the  bore  is  most  com- 
monly used,  but  good  practice  allows  considerable  deviation. 
Throughout,  all  data  are  given  for  well-designed  and  well- 
constructed  machinery  and  not  the  cheap  equipment  that  is 
made  to  sell,  not  to  use. 


66 


BASIC   COSTS 


TABLE  Q 

STEAM  CONSUMPTION 
SIMPLE  HIGH  SPEED  ENGINES  —  NONCONDENSING 


STROKE 

LOAD 

PRESSURE 

8o-go#   " 

IOO-IIO# 

I20-I3O# 

^ 

10" 

| 

30.0 

29.6 

29-3 

29.0 

3 
4 

30.6 

29.8 

29.1 

28.5 

. 

I 

34  -° 

32.2 

30.5 

29.1 

i 

40.0 

38.4 

37-0 

35-6 

l¥ 

30.8 

30.5 

30.2 

30.0 

12" 

1 

29.6 

29.2 

28.9 

28.6 

I 

30.2 

29.4 

28.7 

28.1 

I 

31-4 

30.0 

28.6 

| 

38.7 

37-2 

36.0 

34-8 

ll 

30-3 

30.0 

29.7 

29  5 

14" 

1 

29.1 

28.7 

28.4 

28.1 

f- 

29.7 

28.9 

28.2 

27.6 

' 

i 

32.0 

30.6 

29-3 

28.1 

J 

37-5 

36-1 

35-o 

34-o 

H 

29.8 

29-5 

29.2 

29.0 

16" 

1 

28.5 

28.1 

27.8 

27.5 

i 

29.2 

28.4 

27.7 

27.1 

I 

31.0 

29.8 

28.6 

27-5 

4 

36.2 

35-o 

34-o 

33-° 

if 

29-3 

29.0 

28.7 

28.5 

18" 

1 

27.9 

27-5 

27.2 

26.9 

1 

28.5 

27.8 

27.1 

26.5 

' 

f 

30.0 

29.0 

28.0 

27.0 

i 

35.0 

34-0 

33.0 

32-0 

** 

28.8 

28.5 

28.2 

28.0 

STEAM   CONSUMPTION 


67 


TABLE  10 

STEAM  CONSUMPTION 

HIGH  SPEED  TANDEM  COMPOUND  ENGINES 


STROKE 

LOAD 

PRESSURE 

ioo-no# 

I20-I3O# 

I40-iso# 

ioo-no# 

120-130! 

I40-iso# 

Condensing  26"  vacuum 

Noncondensing 

10-12'' 

* 

19-3 

18-5 

17.7 

25-5 

24.7 

24.0 

f 

20.1 

19-3 

18.6 

28.0 

27.2 

26.4 

1 

22.0 

21.0 

20.4 

32.6 

31-6 

30.6 

i 

26.5 

26.O 

25-5 

44.0 

43-0 

42.0 

i* 

20.8 

19.6 

19.0 

27.0 

25-7 

25-5 

14" 

1 

18.8 

18.0 

17.2 

24.8 

24.0 

23-3 

f 

19-5 

18.7 

18.0 

27.0 

26.2 

25-5 

I 

21.2 

20.4 

19.7 

32.0 

31.0 

30.0 

1 

26.0 

25-5 

25-0 

43-6 

42.6 

41.6 

ii 

19.9 

18.8 

18.2 

26.0 

25.0 

24-5 

16" 

1 

18.2 

17-4 

16.7 

24.0 

23.2 

22.5 

t 

18.9 

18.1 

17-4 

26.0 

25-2 

24-5 

1 

20.4 

19.8 

19.0 

3i-5 

30-5 

29-5 

4" 

25-5 

25.0 

24-5 

43-3 

42-3 

4i-3 

ii 

I9.I 

18.3 

17-5 

25.0 

24.2 

23-5 

18' 

1 

I7-S 

16.7 

16.0 

23.0 

22.2 

21-5 

f 

18.2 

17.4 

16.7 

25.0 

24.2 

23-5 

i 

19-5 

19.2 

18.3 

31.0 

30.0 

29.0 

^ 

25.0 

24-5 

24.0 

43-o 

42.0 

41.0 

ij 

18.3 

17-5 

16.8 

24.0 

23-4 

22.5 

68 


BASIC   COSTS 


TABLE   ii 

STEAM  CONSUMPTION 

SIMPLE  NONCONDENSING  FOUR  VALVE  ENGINES 


STROKE 

LOAD 

PRESSURE 

ioo-s# 

uo-S# 

I20-S# 

I30-S# 

I40-S# 

ISO-S# 

i6o-5# 

18" 

| 

26.0 

25-7 

25-4 

25.2 

24.9 

24.7 

24-5 

1 

27.0 

26.3 

25-6 

25.0 

24-3 

23-8 

23-3 

I 

29.7 

28.5 

27-5 

26.5 

25-5 

24.7 

23-9 

1 
4 

35-5 

34-5 

33-6 

32.7 

31-7 

31.0 

30-3 

ii 

27.1 

26.8 

26.5 

26.2 

26.0 

25-7 

25-5 

21" 

1 

25-8 

25-5 

25-2 

25.0 

24.7 

24.5 

24-3 

i 

4 

26.8 

26.1 

25-4 

24.8 

24.1 

23-6 

23.1 

3 
4 

29-5 

28.3 

27-3 

26.3 

25-3 

24-5 

23-7 

i 

35-o 

34-0 

33-i 

32.2 

31.2 

30.5 

29.8 

if 

26.9 

26.6 

26.3 

26.O 

25-8 

25-5 

25-3 

24" 

1 

25.6 

25-3 

25.0 

24.8 

24-5 

24-3 

24.1 

1 

26.6 

25-9 

25.2 

24.6 

23-9 

23.4 

22.9 

I 

29-3 

28.1 

27.1 

26.1 

25-1 

24-3 

23-5 

i 

4 

34-5 

33-5 

32  .6 

31-7 

30-7 

30.0 

29-3 

ii 

26.7 

26.4 

26.1 

25-8 

25-6 

25-3 

25.1 

27" 

1 

25-4 

25-1 

24.8 

24.6 

24-3 

24.1 

23-9 

i 

4 

26.4 

25-7 

25.0 

24.4 

23-7 

23-2 

22.7 

1 

29.  i 

27.9 

26.9 

25-9 

24.9 

24.1 

23-3 

i 

34-0 

33-o 

32.1 

31.2 

30.2 

29-5 

28.8 

I* 

26.5 

26.2 

25-9 

25-6 

25-4 

25.1 

24.9 

30" 

1 

25.2 

24.9 

24.6 

24.4 

24.1 

23-9 

23-7 

f 

26.2 

25-5 

24.8 

24.2 

23-5 

23.0 

22.5 

1 

28.9 

27-7 

26.7 

25-7 

24.7 

23-9 

23.1 

i 

33-5 

32.5 

31-6 

30-7 

29.7 

29.0 

28.3 

ii 

26.3 

26.0 

25-7 

25-4 

25.2 

24-9 

24.7 

36" 

1 

25.0 

24.7 

24.4 

24.2 

23-9 

23-7 

23-5 

f 

26.0 

25-3 

24.6 

24.0 

23-3 

22.8 

22.3 

I 

28.7 

27-5 

26.5 

25-5 

24-5 

23-7 

22.9 

i 

33-0 

32-0 

3i-i 

30.2 

29.2 

28.5 

27.8 

if 

26.1 

25-8 

25-5 

25.2 

25.0 

24.7 

24-5 

STEAM  CONSUMPTION 


69 


TABLE   12 
STEAM  CONSUMPTION 

CROSS  AND  TANDEM  COMPOUND  ENGINES 


PRESSURE 

STROKE 

LOAD 

IOO-IIO# 

I2O-I3O# 

140-1  so# 

IOO-IIO# 

I20-I30# 

I40-iso# 

Condensing  26"  vacuum 

Noncondensing 

18" 

1 

15-6 

15-15 

14.8 

24.0 

23.0 

22.  0 

1 

16.75 

16.2 

15-5 

24-75 

22.7 

22.7 

i 

17-5 

17-3 

17.2 

26.0 

24.0 

27.0 

1 

4 

20.5 

20.1 

19.8 

41-5 

41  .0 

40-5 

Ij 

16.2 

15-8 

15-5 

25-0 

24.0 

22.  0 

21" 

f 

15-45 

15.0 

I4-65 

23-5 

22.5 

21-5 

4 

16.5 

16.0 

15-4 

24.2 

22.6 

21.6 

1 

17.5 

17-4 

17-3 

26.4 

26.8 

27.1 

1 

22.0 

21.0 

20.3 

41.2 

41.0 

40.7 

I* 

16.1 

15-7 

14.9 

24-5 

23-4 

21.8 

24" 

1 

15-35 

14.9 

14-55 

23.0 

22.0 

21.0 

1 

16.2 

15-7 

15-2 

23-9 

22.6 

21.4 

t 

17.6 

17-5 

17.4 

26.8 

27.0 

27.2 

i 

24.0 

22.5 

20.8 

40.9 

40.9 

40.9 

ii 

16.0 

15-7 

15-3 

24.0 

23.0 

21.6 

27" 

4 

4 

15-1 

14-75 

14.4 

22.5 

21-5 

20.5 

1 

16.0 

15-6 

15-1 

23-3 

22.3 

21.3 

1 

17.7 

I7.6 

17.6 

27.2 

27.3 

27-3 

1 
4 

26.0 

24.0 

21-5 

40.6 

40.9 

4I.I 

J¥ 

15-8 

15-5 

15-1 

23-3 

22.3 

21.4 

30" 

4 

15.0 

14.6 

14-25 

22.  0 

21.0 

20.0 

a 

4 

15-8 

15-4 

14.9 

22.6 

21.8 

21.1 

1 

17.7 

17.7 

17.6 

27.6 

27.4 

27.4 

i 

28.0 

25-0 

22.  0 

40-3 

40.8 

41-3 

!¥ 

15-7 

15-3 

15-0 

22.6 

21.8 

21.2 

36" 

1 

14.6 

14.4 

14.0 

21-5 

20.0 

I9.O 

f 

15-6 

15-2 

I4.8 

22.  O 

21-5 

21.0 

1 

17.8 

I7.8 

I7.8 

28.0 

27-5 

27-5 

i 

30-7 

26.6 

22.6 

40.0 

40.7 

41-5 

11 

15-5 

15-2 

14.8 

22.  0 

21  .2 

21.  0 

BASIC   COSTS 


TABLE   13 

STEAM  CONSUMPTION 

HORIZONTAL  SINGLE  CYLINDER  LENTZ  ENGINES 
Referred  to  i.h.p.  at  normal  load  and  dry  saturated  steam 


STROKE 

PRESSURE 

85# 

IOO# 

«5# 

I2S# 

I40# 

ISO# 

i6o# 

Noncondensing 

1  8" 

24-6 
24.1 
23-8 
23-4 
23-0 
22.6 
22.2 

23.6 
23.1 

22.8 
22.4 
22.0 
21.6 
21  .  2 

22.7 
22.2 
21  .9 

21-5 
21.  I 

20.7 
20.3 

22  .1 
21  .6 

21.3 
2O.9 
20.5 
2O.  I 
19.7 

21-3 
20.8 

20.5 

20.  1 

19.7 

19-3 
18.9 

20.8 

20.3 

20.  o 
19.6 

19.2 
18.8 
18.4 

20.3 
19.8 

I9-S 
I9.I 
I8.7 
I8.3 
17.9 

14,  15,  16  and  17  X  21"  
18,  19,  20  and  21  x  21"  
24".  . 

27" 

<o".  . 

33,  36,  42".  . 

jg*  

Condensing 

20.3 
19.9 
19.7 
19.4 
19.1 
18.8 
I8.5 

19.7 
19.2 
18.9 

18.6 

18.2 
18.0 
17.7 

I8.7 
l8.4 

18.1 
17-8 

17-5 
17.2 
16.9 

18.2 

17.9 

17.6 

17-3 
17.0 

16.7 
16.4 

17-5 
17.2 

16.9 
16.6 

16.3 
16.0 

15-7 

17.1 
16.8 

16.5 
16.2 

15.9 

15.6 

15-3 

I6.7 
16.4 

16.1 

15-8 
i5-S 
15-2 
14.9 

14,  15,  16  and  17  x  21"  

18,  19  and  21  x  21"  
24"  . 

27".  . 

30".  . 

33,  36,  42"  

For  50  degrees  superheat  reduce  consumption   8%,  100  D  superheat   12%, 
150  D  17%,  200  D  20%. 

At  fractional  loads,  increase  steam  consumption  as  follows: 

LOAD  l|  f  f  I  i 

Saturated  steam 5%  o%  4%  9%  25% 

Superheated  steam 5%  o%  3%  7%  18% 


STEAM  CONSUMPTION 


TABLE  14 
STEAM  CONSUMPTION 

HORIZONTAL  CROSS  AND  TANDEM  COMPOUND  ENGINES 

Referred  to  i.h.p.,  normal  loads  and  dry  saturated  steam 


PRESSURE 

STROKE 

ioo# 

"5# 

I25# 

I40# 

iSO# 

i6o# 

I7S# 

IQO# 

200# 

Noncondensing 

18" 

22.1 

21  .  2 

20.  6 

19.8 

19-3 

18.8 

18.1 

17-5 

I7.I 

21" 

21.8 

2O.9 

20.3 

IQ-5 

19.0 

18.5 

17.8 

17.2 

16.8 

24" 

21.5 

20.6 

20.0 

19.2 

I8.7 

18.2 

17-5 

l6.9 

16.5 

27" 

21  .  2 

20.3 

19.7 

18.9 

18.4 

17.9 

17.2 

16.6 

16.2 

30" 

20-9 

20.  o 

19.4 

18.6 

18.1 

17-6 

16.9 

16.3 

15-9 

33,  36,  42* 

20.  6 

19.7 

I9.I 

18.3 

18.8 

17-3 

16.6 

16.0 

iS-6 

Condensing 

18" 

16.4 

15.8 

15-35 

14.8 

14-45 

14,1 

i3  6 

13-2 

12.9 

21* 

16.1 

15-5 

I5-05 

14-5 

14-15 

13-8 

13  3 

12.9 

12.6 

24" 

15-8 

15-2 

14-75 

14.2 

13-85 

13-5 

13.0 

12.6 

12.3 

27" 

15-5 

14.9 

14-45 

13-9 

13-55 

13-2 

12.7 

12.3 

12.0 

30" 

15-2 

14.6 

14-15 

13-6 

13-25 

12.9 

12.4 

12.0 

II.7 

33,  36,  42" 

14.9 

14-3 

13-85 

13-3 

12.95 

12.6 

12.  I 

II.7 

II.4 

For  50  degrees  superheat  reduce  consumption  8%,  for  100  D  12%,  for  150  D 
17%,  and  for  200  D  20%. 

Increase  steam  consumption  for  fractional  loads  as  follows: 

LOAD  l|  |  f  f  I 

Saturated  steam 6%  o%  3%  7%  20% 

Superheated  steam 6%  o%  2^%  6%  15% 

60.  The  Steam  Consumption  of  simple  low  speed  condensing 
engines  is  18.5%  greater  than  that  of  compound  condensing 
engines. 

The  Steam  Consumption  of  simple  low  speed  noncondensing 
engines  is  42%  greater  than  that  of  compound  condensing 
engines. 


72  BASIC  COSTS 

So  also  the  Steam  Consumption  of  simple  high-speed  non- 
condensing  engines  is  64%  greater  than  that  of  compound 
condensing  engines. 

The  foregoing  steam  consumption  of  compound  condensing 
engines  corresponds  to  the  following  thermal  efficiencies,  based 
on  an  initial  normal  water  temperature  of  52°  F. 

TABLE   15 

THERMAL  EFFICIENCIES 
COMPOUND  CONDENSING  ENGINES 

I.h.p 50  60  70  80  loo  125  150  200       250 

Efficiency  % 10.8  n.i  11.4  11.6  12.0  12.4  12.6  13.1      13.3 

I.h.p 300  400  500  600  700  800  900  1000 

Efficiency  % 13.5  13.9  14.2  14.4  14.5  14.6  14.6  14.7 

This  corresponds  to  the  equation 

16.6  VM 

y  =  -TTT. — :» (25) 


where  y  =  the  efficiency  in  per  cent, 
and  M    =  the  size  in  i.h.p. 

61.  With  superheated  steam,  the  steam  consumption  is 
decreased.  For  50°  F.  superheat,  the  steam  consumption  is 
reduced  by  4%,  for  100°  superheat  12%,  for  150°  superheat 
17%,  and  for  200°  superheat  20  %.  This  corresponds  to  increases 
in  efficiency  of  4%  for  50°  superheat,  8.4%  for  100°,  12.6%  for 
150°,  and  an  increase  in  efficiency  of  16.6%  for  200°  F.  super- 
heat, based  on  the  efficiencies  obtained  for  dry  saturated  steam. 
This  shows  that  the  efficiency  increases  in  direct  proportion  to 
the  increase  in  superheat,  according  to  the  equation: 

i  —  0.0835 (26) 

where  i  =  the  increase  in  efficiency  in  per  cent, 
s  =  the  degrees  superheat. 

The  variation  of  consumption  of  dry  saturated  steam  with 
variations  of  initial  steam  pressure  is: 

(a)  For  compound  condensing  engines 

W  =  41.6  —  0.074^, (27) 

where  W  =  the  pounds  of  steam  used, 
and       E  =  the  temperature  of  the  steam. 


FRACTIONAL  LOADS  73 

(b)  For  low  speed  simple  noncondensing  engines 

W  =  57-4  -  o.iE (28) 

The  corresponding  efficiencies  are: 

(a)  For  compound  engines 

y  =  o.o675(£  -  145) (29) 

(b)  For  simple  low  speed  noncondensing  engines  we  have 
approximately 

y  =  o.o466(E  —  120) (30) 

These  formulae  are  obtained  from  the  data  above  given. 
While,  like  all  the  above  data,  these  efficiencies  are  based  on 
factory  guarantees,  we  do  not  consider  them  entirely  satisfac- 
tory. They  are,  however,  sufficiently  accurate  for  ordinary 
practical  use. 

62.  Equipment  is  very  rarely  operated  at  full  load.  Nor- 
mally it  is  always  run  at  some  fraction  of  the  full  load.  Far 
more  important  than  full  load  efficiency  are  the  efficiencies  at 
fractional  loads,  for  upon  these,  rather  than  those  at  full  load, 
will  the  economy  of  operation  depend.  We  therefore  give  the 
variation  in  steam  consumption  at  fractional  loads.  These 
are  as  follows: 

(a)  For  compound  condensing  engines  the  per  cent  increase 
in  steam  consumption  at  various  per  cents  of  full  load  is  as 
follows : 

TABLE   16 

Fraction  of  full  load  (F) 0.25  0.50  0.75  i.oo  1.25 

Per  cent  full  load,  dry  sat.  steam  (PSat.)    122  108  101  100  106 
Steam  consumption,  superheated  steam 

(•Ssup.) 117  106  loo  zoo  105 

The  equations  for  these  values  are 

i 


sat. 


57.6  -  53F(F  -  1.79) 
and 


sup. 


66.8  -  42.8F(F  -  1.78) 


(32) 


where  (F)  is  the  fraction  (per  cent)  of  full  load.     And  the 
corresponding  per  cents  of  full  efficiency  are 

y»t    =  57-6  -  S3F(F  -  1.79) (33) 

and  ;ysup.    =  66.8  -  42.8F(F  -  1.78) (34) 


74  BASIC  COSTS 

For  low  speed  simple  engines  the  values  are: 

TABLE   17 

Fraction  of  full  load  (F) 0.25  0.50  0.75  i .  oo  1.25 

Per  cent  full  load,  dry  sat.  steam  (PSat.) 125  109  101  100  107 

Steam  consumption,  superheated  steam  (Psup.)  ..     117  106  100  100  105 


The  equations  for  these  values: 


sat. 


52  -  6oF(F  -  1.8)' 


p       _  t 

66.8  -  42.8F(F  -  1.78) 


(35) 

•    (36) 


The  corresponding  per  cents  of  full  load  efficiencies  are: 
y»t  =  52  -6oF(F  -  1.8) (37) 

and  ;ysup.  =  66.8  -  42.SF(F  -  1.78), (38) 

the  latter  being  the  same  as  for  compound  engines. 


0    100    200    300    400    500    600    700    800    900   1000 

I.  H.  P, 

FIG.  2.  —  Steam  consumption  and  attendance  costs  of  compound 
condensing  steam  engines. 

63.  The  attendance  cost  of  steam  engines  alone  is  of  com- 
paratively little  value,  as  they  must  always  be  run  as  part  of  a 
system.  Based ,  however,  on  the  mean  of  the  most  authoritative 
data,  this  is  about  as  follows: 


BOILERS  75 

TABLE   18 

1906-10  PRICES 

Size,  i.h.p  .........       10        20          30        40  50         60          75          100 

Attend.,  cts.  per  i.h.p.    0.47    0.38      0.30    0.275      °-26    0.248    0.235      0-22 

Size,  i.h.p  .........     150        200        300      400        500        600        800        1000 

Attend.,  cts.  per  i.h.p.  0.192    0.172    0.1520.137    0.127     °-12     0.108     o.ioo 

Size,  i.h.p  ......  ".  ........................     1500  2000 

Attend.,  cts.  per  i.h.p  .....................  0.0885  0.08 

This  corresponds  approximately  to  the  equation 

.......  (39) 


where  A  —  attendance  cost  in  cents  per  i.h.p.h., 
and    M  =  the  i.h.p.  size  of  the  engines. 

The  cost  of  such  incidentals  as  oil,  waste,  and  supplies  is 
about  one-fourth  of  one  per  cent  as  great  as  the  attendance  cost. 

64.  Boilers.  —  With  present  engineering  development,  there 
is  no  apparatus  more  used  than  the  steam  boiler.  Under  such 
conditions,  it  would  be  only  natural  to  assume  that  the  worth 
of  the  investment  for  such  equipment  would  have  been  thor- 
oughly investigated.  However,  quite  the  opposite  is  the 
case.  It  is  true  that  an  enormous  number  of  boiler  tests  have 
been  made.  But  the  results  of  these  tests  are  so  erratic  that 
practically  no  confidence  can  be  placed  in  them  as  far  as  the 
deduction  of  any  general  law  of  variation  of  efficiency  with 
size  is  concerned.  Measurements  in  heat  are  extremely  diffi- 
cult, in  fact  so  much  so  that  the  difficulty  can  hardly  be  exag- 
gerated. The  same  discrepancy  appears  in  tests  at  fractional 
loads. 

Amongst  the  great  mass  of  tests  made,  a  few  stand  out  as 
worthy  of  our  entire  confidence,  though  they  are  unfortunately 
too  few  to  yield  conclusive  evidence.  The  most  noteworthy 
of  these  is  the  test  of  the  boilers  of  the  Detroit  Edison  Com- 
pany, made  by  a  corps  of  experts  under  D.  S.  Jacobus,  the  results 
of  which  are  published  in  Vol.  33  of  the  American  Society  of 
Mechanical  Engineers.  The  boilers  had  23,650  sq.  ft.  of  heat- 
ing surface  each.  They  showed  79.3%  efficiency  at  rated  load 


76 


BASIC   COSTS 


and  80.5%  at  70%  of  rated  load.  With  increase  of  load,  there 
was  a  steady  fall  in  efficiency  as  far  as  the  tests  were  conducted. 
Tests  were  also  made  of  the  same  type  of  boiler  (Stirling)  of 
6400  sq.  ft.  of  heating  surface,  by  J.  A.  Hunter.  They  showed 
78%  efficiency  at  rated  load,  79.8%  efficiency  at  70%  of  rated 
load  and  66%  efficiency  at  200%  of  rated  load.  These  tests 
are  well  in  consonance  with  the  Jacobus  tests.  They  are 
reported  in  full  in  Vol.  31  of  the  American  Society  of  Mechanical 
Engineers.  Further  data  of  unusual  value  on  boilers  are  given 


ncy 

g 


cent 

§ 


50 


100         200        300         400         500         600         700        800         900      1000 
Bo.  H.P. 

FIG.  3.  —  Variation  in  boiler  efficiency,  with  size. 


by  A.  D.  Pratt  in  a  paper  that  he  presented  before  the  Inter- 
national Engineering  Congress  at  San  Francisco,  California,  in 


Taken  as  a  whole,  most  manufacturers  of  boiler  sell  heating 
surface  rather  than  boiler  service.  The  results  obtained  from 
boilers  depends  in  a  very  large  measure  on  how  the  boiler  is 
installed,  the  nature  of  the  fuel  used,  and  the  methods  of  han- 
dling them.  Good  results  can  only  be  obtained  where  the 
entire  boiler  installation  is  made  in  accordance  with  the  direc- 
tion of  boiler  experts. 

65.  In  the  table  below  we  give  the  factory  prices  of  one  make 
of  boilers.  These  are  at  times  somewhat  erratic,  although  they 
are  as  a  whole  fairly  well  in  proportion.  To  bring  this  out  clearly 
we  have  added  the  last  column  of  adjusted  prices.  Some  of 
the  difference  between  the  factory  and  the  adjusted  prices  is  no 
doubt  due  to  error  in  the  cost  calculations  at  the  factory,  but 
the  greater  part  is  due  to  commercial  exigencies,  such  as  the 


BOILERS 


77 


adaptation  of  the  boiler  dimensions  to  the  use  of  stock  sizes  of 
sheets  and  tubes.  The  result  of  this  is  that  the  actual  and 
rated  sizes  of  boilers  are  not  always  the  same.  Allowing  for 
this,  the  agreement  between  the  factory  and  adjusted  prices  is 

excellent. 

TABLE   19 

STATIONARY  RETURN  TUBULAR  BOILERS  WITH  FULL  FRONT  SETTING 
1906-10  PRICES 


Bo.h.p. 

Size 

Cost  factory, 
total 

Per  bo.h.p. 

Adjusted  cost 
per  bo.h.p. 

IO 

30"  x   7' 

$166.00 

$16.60 

$17.25 

12 

30"   X     8' 

176.00 

14.67 

I5-46 

15 

36"  X    8' 

207.00 

13.80 

13-66 

20 
2O 

36*  X  10' 

42"  X    8' 

236.00 
253.00 

ii.Sp  \ 
12.6s/ 

11.87 

25 

36"    X  12' 

253.00 

10.12  1 

10.  80 

25 

42"  X  10' 

292.00 

II.68J 

30 

42"  X  12' 

313.00 

10.43! 

_jT 

IO.  OO 

30 

44"  X  10 

310.00 

10.33] 

35 

44"  X  12' 

340.00 

9.72 

9-57 

40 
40 

44"  X  14' 
48"  X  12' 

397-50 

9-37J 
9-94J 

9.17 

45 

48"  X  14' 

435-oo 

9.67 

8.90 

50 

48"  x  1  6' 

472.00 

9-44 

8f\r 

50 

54"  X  12' 

465.00 

9-30 

•"5 

60 

54"   X  14' 

508.00 

8-47 

8. 

60 

60"  X  12' 

550.00 

9.17 

.    29 

65 

54"  X  i/ 

530.00 

8.18 

8.15 

70 
70 

54"  X  16' 
60"  X  14' 

555-oo 
600.00 

7-93\ 
8-57J 

8.04 

75 

60"  X  15' 

622.00 

8-33 

7-94 

80 

60"  X  16' 

650  .  oo 

8.15 

7-84 

85 

66"  x  14' 

720.00 

8.50 

7.76 

90 

66"  x  15' 

730.00 

8.  ii 

7.70 

100 

66"  x  1  6' 

770.00 

7.70 

7-57 

no 

66"  x  18' 

820.00 

7-45 

7.48 

125 

72"  x  16' 

920.00 

7-36 

7-36 

150 

72"  X  18' 

1000.00 

6.67 

7.22 

The  equation  for  the  adjusted  prices  is 

c -=**  +  *» 

where  C  =  cost  in  dollars  per  bo.h.p. 
and     M  =  the  size  in  bo.h.p. 


(40) 


BASIC   COSTS 


TABLE  20 

WATER-TUBE  BOILERS  WITH  FULL  FRONT  SETTING 
PRICES,  1906-10 


Size,  bo.h.p 


100 


125 


175 


200 


250 


Price  per  bo.h.p $13.60      $12.70      $12.10      $11.67      $u-35      $10.90 

Size,  bo.h.p 300  350  400  450  500  600 

Price  per  bo.h.p $10.60      $10.39       $10.22       $10.10      $10.00        $9.85 

The  equation  for  these  values  is 

c  =  4s° 


M 


+  9.10. 


Fernold  and  Orrok,  "  Engineering  of  Power  Plants/'  give 


(42) 


and  the  cost  of  setting  as 

Cs  =  140  +  2M (43) 

66.    Cost  of  Building.  —  These  same  authorities,  Fernold  and 
Orrok,  give  average  data  on  the  cost  of  buildings,  and  the  divi- 


Cost  per  Bo.  H.P.  & 

s  a  s  g 

5 

^v 

'  

—         •. 

^  !•, 

•= 

_W£ 

-*            m 

ter  1 

~         

•ube 

gpile 

'8 

••      • 

—  ^_ 

——  . 

^> 

'^*. 

—  «*. 

—  —  —  . 

Ret 

=9MM 

urn  ' 

—  — 

ubul 

jrB« 

—  — 

lers 

^_ 

)           25          50          75         100         125         150        175        200        225       25C 
Bo.  H.P. 

FIG.  4.  —  Variation  in  prices  of  boiler  with  size  and  type. 

sion  of  costs  in  power  plants  in  good  practical  form,  some  of 
which  is  given  below, 


BUILDINGS  79 

TABLE  21 
COST  OF  BUILDINGS,  1906-10 

Type                                                          Cost  in  dollars  Cost  in  cents 

per  sq.  ft.  per  cu.  ft. 

floor  surface  of  contents 

Mill  construction o. 80  to  i .  10  6 . 5  to  8.5 

Fireproof  stores,  factories  and  warehouses  with 

brick,  concrete,  stone  and  steel  construction.  .  .  2.00  to  3.00  14.0  to  25.0 
Concrete  and  reiniorced  concrete  shops,  factories 

and  warehouses i .  25  to  i .  75  8.0  to  16.0 

Plain  power  houses,  with  concrete  floors,  brick  and 

steel  super-structure 2 .  oo  to  2 . 75  9 .  o  to  12 .  o 

Power  houses  under  city  conditions  with  superior 

architecture 3.00  to  4. 50  15.01030.0 

Exact  data  cannot  possibly  be  given,  because  the  cost  of 
buildings  varies  enormously  with  local  conditions.  These  must 
always  be  determined  by  the  engineer  before  reaching  any  final 
decision. 

TABLE   22 

COST  OF  STEAM  POWER  PLANT  BUILDINGS 
1906-10 

PER  ENGINE  HORSE  POWER 

Simple  Noncondensing  Engines 

H.p 10           12          15          20          30          40  50          75 

Boiler  house $37. 15  $33.00  $28.50  $24.50  $20.50  $18.00  $16.00  $13.00 

Engine  house 4.80      4.35       3.90       3.30       2.75       2.50  2.30      2.15 

Coal  bins 20.00     18.00     16.00     13.70     n.oo      9.80  8.30      6.00 

Simple  Condensing  Engines 

H.p 10          12         15         20         30         40         50  75       100 

Boiler  house $33.70  $29.60  $26.20  $21.60  $18.20  $16.00  $14.80  $11.30  $9.70 

Engine  house. ...   14.40     12.60     10.90       8.60       7.75       6.40      5.35  4.00    4.30 

Coal  bins 19.00     17.90     15.80     13.60    11.00      8.70      8.50  6.30    5.70 

Compounding  Condensing  Engines 

H.p 100         200            300        400       500      600  700      800 

Boiler  house 1  $                            f  11.20    8.00    6.40    5.70  5.35     5.00 

Engine  house j                               \ii.2o    9.35     8.50    7.20  6.30    5.60 

Coal  bins 5.70      4.00          3.10    2.60    2.40    2.25  2.10    2.05 

H.p 900                       1000                      1500  2000 

Boiler  house $4-7°                   $4-55                   $4-i°  $3-95 

Engine  house 5.35                     5.00                     4.75  4.55 

Coal  bins 1.95                      1.80                      1.75  1.66 


8o  BASIC   COSTS 

The  total  cost  of  engine  and  boiler  house  and  coal  bins  for 
simple  noncondensing  engines  is  given  by  the  equation 

Ci  =  63.5  -  J  \/2ooM  -  M2  -  1900.    .    .  (44) 
For  simple  condensing,  this  is 


C2  =  65.0  —  J  V2ooM  —  M2  —  1900,   .    .    (45) 
while    for    compound    condensing    engines    the    approximate 
equation  is 


.      C.-  +  8.2..    .    .    .    .    .    (46) 

67.   The  cost  of  complete  steam  installations  is  given  as 
follows  : 

TABLE   23 

H.p  ................    5000  10,000        20,000       30,000       40,000       50,000 

Cost  per  h.p  ........  $120.00      $100.00     $90.00      $80.00      $75.00     $70.00 

These  costs  are  found  to  be  divided  about  as  follows: 


TABLE   24 

Percentage  of  total 

Buildings,  real  estate  and  excavations 14.6 

Turbines  and  generators 23 . 5 

Condensers 5.7 

Boilers,  superheaters,  stokers  and  stacks 28. 7 

Bunkers  and  conveyors 4.8 

Boiler,  feed  and  service  pumps i .  o 

Feed  water  heaters 1.6 

Switchboard  and  wiring , 3.5 

Exciters 2.1 

Foundations  —  machinery i .  i 

Piping  and  conduit 7.1 

Crane 1.5 

Superintending  and  engineering 4.8 

68.  Centrifugal  Pumps.  —  The  variation  in  price  of  one 
certain  make  of  centrifugal  pumps  with  size  and  type  is  given 
below. 


BELTED   CENTRIFUGAL  PUMPS 


81 


TABLE  25 

Low  HEAD  HORIZONTAL  BELTED  CENTRIFUGAL  PUMPS 
1910-12 


Size 
inches 

Normal  capacity, 
g.p-m. 

Cost  of  pump 

Cost  per 
1000  g.p.m. 

Adjusted 
costs 

I 

20 

$27.00 

$1350.00 

if 

50 

33-oo 

66o.OO 



2 

IOO 

42.00 

42O.OO 

2\ 

150 

51.00 

34O.OO 

m  * 

3 

225 

60.00 

266.OO 

3^ 

300 

69.00 

230.00 



4 

400 

75.00 

187  .50 

• 

< 

700 

IOO.OO 

142.86 

O 

6 

900 

I  20.00 

133-33 



7 

I2OO 

150.00 

125.00 



8 

I600 

180.00 

112.50 

$180.70 

10 

3000 

240.00 

So.OO 

253-10 

12 

4500 

300.00 

66.67 

330-75 

14 

6OOO 

425.00 

70.83 

408  .  2O 

15 

7000 

475.00 

67.86 

459-90 

16 

8000 

525-00 

65.62 

511.60 

18 

IO,OOO 

900.00 

9O.OO 

615.00 

20 

I4,OOO 

1000.00 

71.40 

821.80 

24 

20,000 

I2OO.OO 

6o.OO 

1132.00 

30 

30,000 

1550.00 

5I-67 

1649.00 

36 

4O,OOO 

2400  .  oo 

6o.OO 

2l66.OO 

40 

50,000 

2700.00 

54-oo 

2683.00 

44 

60,000 

3200.00 

53-33 

32OO.OO 

The  factory  costs  are  somewhat  erratic  due  to  the  lack  of 
knowledge  of  exact  costs  on  the  part  of  the  manufacturer. 
The  prices  on  the  larger  sizes  are  no  doubt  based  on  "  esti- 
mates." The  equation  based  on  the  adjusted  values,  i.e.,  those 
that  have  been  brought  into  consonance  with  themselves,  is 


(47) 


where     C  —  cost  in  dollars  per  1000  g.p.rn.  of  capacity, 
and       M  =  the  capacity  in  zooo's  of  g.p.m. 


BASIC  COSTS 


0  4  8  12  16          20          24  28          32          36          40 

Size  of  Pump  in  Inches  of  Discharge  Nozzle 

FIG.  5.  —  Variation  in  price  of  centrifugal  pumps,  with  size  and  type. 


TABLE  26 

HIGH  HEAD  HORIZONTAL  BELTED  CENTRIFUGAL  PUMPS 
1910-12 


Size 
inches 

Normal  capacity, 
g.p.m. 

Cost,  factory 

Cost  per 
1000  g.p.m. 

Adjusted 
costs 

II 

SO 

$67.20 

$1344.00 

$77-625 

2 

IOO 

84.00 

840.00 

83-75 

*\ 

150 

94.80 

632.00 

90-375 

3 

225 

112.  2O 

500.00 

99.94 

4 

400 

I34-40 

338.00 

122.25 

5 

700 

150.00 

214.30 

160.50 

6 

900 

198.00 

220.00 

186.00 

7 

1200 

24O.OO 

2OO.OO 

224.25 

8 

l6oO 

294.00 

183.75 

275.25 

10 

2500 

390.00 

156.00 

390.00 

12 

4OOO 

522.00 

130.50 

581.25 

The  equation  based  on  the  adjusted  values  is 
C=  ^TF5  +  $I27-5°-     • 


(48) 


DIRECT-CONNECTED   CENTRIFUGAL  PUMPS        83 


TABLE  27 

DIRECT-CONNECTED  HORIZONTAL  CENTRIFUGAL  PUMPS  LESS  MOTOR 
1910-12 

25  Ft.  Head 


H.p. 

R.p.m. 

Size 
inches 

Capacity 
g.p.m. 

Cost 
factory 

Cost  per 
looo  g.p.m. 

i 

1800 

I 

IS 

$70.00 

$4666.67 

I 

1800 

3 

70 

IOO.OO 

1428.00 

2 

1800 

3* 

130 

125.00 

960.00 

3 

1800 

4 

240 

140.00 

580.00 

5 

1800 

5 

420 

170.00 

405.00 

7i 

1200 

5 

610 

230.00 

376.00 

7^ 

I2OO 

6 

650 

310.00 

476.00 

10 

I2OO 

5 

820 

240.00 

294.00 

10 

I2OO 

8 

IOOO 

385.00 

385-00 

15 

QOO 

7 

1300 

310.00 

237.00 

20 

I800 

7 

1350 

330.00 

245.00 

20 

720 

10 

2300 

500.00 

2l8.OO 

25 

720 

10 

2500 

530.00 

212.  OO 

30 

1800 

8 

2000 

385-00 

I92.OO 

30 

900 

10 

3200 

500.00 

156.00 

35 

600 

12 

3500 

630.00 

iSo.OO 

50 

9OO 

12 

4800 

675.00 

144.00 

50 

600 

12 

4800 

750.00 

156.00 

84 


BASIC   COSTS 


TABLE   28 

DIRECT-CONNECTED  HORIZONTAL  CENTRIFUGAL  PUMPS  WITHOUT  MOTOR 

1910-12 

50  Ft.  Head 


H.p. 

R.p.m. 

Size 
inches 

G.p.m. 

Cost 

Cost  per 
looo  g.p.m. 

2 

1800 

I 

25 

$80.00 

$3200.00 

2 

1800 

2 

50 

IOO.OO 

2000.00 

3 

1800 

H 

100 

115.00 

1150.00 

5 

3600 

3 

200 

135-00 

675.00 

5 

1800 

2^ 

2OO 

145.00 

725.00 

7i 

1800 

4 

350 

180.00 

512.00 

10 

1800 

4 

500 

185.00 

370.00 

i5 

1800 

4 

700 

210.00 

300.00 

20 

I2OO 

6 

IOOO 

27O.OO 

270.00 

20 

I800 

7 

IIOO 

330.00 

300.00 

25 

QOO 

7 

1250 

400.00 

320.00 

30 

I800 

8 

1500 

385-00 

260.00 

35 

900 

8 

2OOO 

420.00 

210.00 

40 

900 

8 

22OO 

420.00 

191.00 

40 

900 

12 

2600 

930.00 

357-oo 

So 

900 

IO 

25OO 

530.00 

212.  OO 

75 

900 

IO 

3900 

550.00 

141  .OO 

100 

900 

12 

5500 

650  .  oo 

IlS.OO 

125 

600 

15 

7000 

IOOO.OO 

143-00 

150 

720 

14 

775° 

900  .  oo 

n6.co 

DIRECT-CONNECTED   CENTRIFUGAL   PUMPS       85 


TABLE  29 

DIRECT-CONNECTED  HORIZONTAL  CENTRIFUGAL  PUMPS  WITHOUT  MOTOR 

1910-12 

100  Ft.  Head 


H.p. 

R.p.m. 

Size 
inches 

G.p.m. 

Cost 

Cost  per 
1000  g.p.m. 

3 

3600 

I 

43 

$140.00 

$3260.00 

5 

3600 

3 

135 

135-00 

IOOO.OO 

7i 

I800 

2 

125 

150.00 

1  200  .  00 

IS 

3600 

si 

300 

150.00 

500.00 

15 

I800 

2* 

315 

185.00 

588.00 

20 

1800 

4 

470 

210.00 

446.00 

25 

1800 

4 

560 

235.00 

420.00 

30 

I800 

5 

700 

26o.OO 

37l-<» 

35 

1800 

4 

700 

24O.OO 

342.00 

40 

I2OO 

5 

1040 

325.00 

313.00 

So 

1800 

6 

1275 

27O.OO 

212.00 

75 

I2OO 

8 

1800 

42O.OO 

233.00 

75 

1200 

7 

2OOO 

530.00 

265.00 

IOO 

I2OO 

8 

2450 

550.00 

217.00 

IOO 

QOO 

8 

2700 

640.00 

237.00 

150 

1800 

12 

3800 

IOOO.OO 

264.00 

150 

900 

10 

4OOO 

680.00 

170.00 

2OO 

I2OO 

12 

5600 

680.00 

121.  OO 

250 

900 

14 

7000 

1100.00 

157-00 

$3500  r 

3000  1 

& 

'o  9^nnl- 

; 

|  2o00r 

o  9000  1- 

\\ 

si 

_  l^nnL 

\\ 

v 

«  loooN 

\ 

s^ 

5 

\  1 

s 

X* 

^^ 

cnn   \     \, 

^> 

^ 

-<s 

ffhl 

fficie 
•^=1£ 

noy 
tOft. 

Heac 

500^-K 

s>  — 

•^s 

~~Low  Ef 

icien 

yj 

1  r 

0           25          50          75          100         125        150        J75         200        225        25 
H.P.  of  Pump 

FIG.  6.  —  Prices  of  direct-connected  centrifugal  pumps. 


86 


BASIC  COSTS 


TABLE  30 

DIRECT-CONNECTED  HORIZONTAL  CENTRIFUGAL  PUMPS  WITHOUT  MOTOR 

1910-12 

150  Ft.  Head 


H.p. 

R.p.m. 

Size, 
inches 

G.p.m. 

Cost 

Cost  per 
1000  g.p.m.  cap. 

3 

1800 

I 

17 

$230.00 

$13,530.00 

5 

1800 

I 

30 

230.00 

7,670.00 

7* 

1800 

ii 

50 

270.00 

5,400.00 

10 

1800 

2 

90 

270.00 

3,000.00 

IS 

3600 

3* 

150 

150.00 

1,000.00 

20 

1800 

2* 

160 

190.00 

1,190.00 

20 

1800 

4 

300 

430.00 

1,430.00 

25 

1800 

3 

235 

2OO.OO 

.850.00 

35 

1800 

4 

500 

450.00 

900.00 

35 

1800 

5 

550 

575-oo 

1,042.00 

40 

I2OO 

4 

43° 

415.00 

965.00 

50 

I800 

5 

800 

600.00 

750.00 

5° 

I800 

6 

850 

725.00 

850.00 

50 

I2OO 

5 

600 

510.00 

850.00 

75 

I2OO 

6 

1  200 

500.00 

416.00 

75 

1800 

8 

1400 

900.00 

643.00 

IOO 

I2OO 

8 

1500 

55o.oo 

366.00 

100 

I800 

8 

1800 

925-00 

5i3-oo 

VERTICAL   CENTRIFUGAL  PUMPS 


OrtA 

/ 

/^ 

I7f\f\ 

/ 

/ 

x 

Cost  of  Pump  Complete 

•s?^v 

// 

x 

v 

x 

x 

~? 

x 

X 

x 

' 

^. 

^> 

x 

/ 

/^ 

, 

/^ 

•^ 

x^ 

^ 

x 

/ 

? 

^ 

x^ 

x 

x 

x 

? 

/ 

x 

x 

/ 

x 

x 

7 

X 

' 

X 

X 

X 

> 

X 

2^                        456 
Size  of  Pump  in  Inches  of  Discharge  Nozzle.D 

9            1( 
lameter 

FIG.  7.  —  Prices  of  vertical  centrifugal  pumps,  complete. 


TABLE  31 

VERTICAL  CENTRIFUGAL  PUMPS 
1910-12 

Complete  with  frames,  valves,  shafting,  boxes,  pipe,  collars,  etc. 


SIZE, 
INCHES 

NORMAL 
CAPACITY 

PRICE 

With  25  ft.  frame 

With  50  ft.  frame 

With  75  ft.  frame 

2 

100  g.p.m. 

$196.00 

$265.00 

$334-00 

2^ 

150      " 

221  .OO 

293.00 

364.00 

3i 

225      « 

249.00 

332.00 

408.00 

300      " 

281.00 

368.00 

452.00 

4 

400      " 

300.00 

397.00 

482.00 

5 

700      « 

353-oo 

462.00 

561.00 

6 

900      « 

423.00 

550.00 

658.00 

7 

1200 

496.00 

616.00 

733-oo 

8 

1600         " 

605.00 

740.00 

860.00 

88  BASIC   COSTS 

69.  Cost  of  Large  Pumping  Units. — The  following  quotations 
were  made  on  various  types  of  complete  pumping  plant  f  .o.b.  factory, 
between  1908  and  1910. 

1.  A  25,000  g.p.m.  horizontal  pump  for  a  total  head  of  16  ft.,  direct- 
connected  to  a  150  h.p.  257  r.p.m.  a.c.  motor  together  with  primer,  etc., 
and  100  ft.  of  piping.    Price  quoted  $6000. 

2.  Same  as  (i)  above  but  for  20  ft.  head.     Price  quoted  $7500. 

3.  A  25,000  g.p.m.  horizontal  pump  for  a  total  head  of  16  ft.,  direct 
connected  to  a  150  h.p.  horizontal  tandem  compound  condensing  steam 
engine,   complete  with  water  tube  boiler,  jet  condenser,   and   necessary 
piping.     Price  quoted  $12,000. 

4.  Same  as  (3)  but  for  20  ft.  head.     Price  quoted  $15,000. 

6.  One  50,000  g.p.m.  horizontal,  double  suction  pump,  direct-connected 
to  one  300  h.p.  200  r.p.m.  a.c.  motor  for  16  ft.  total  head,  together  with 
loo  ft.  of  piping  and  electric  primer.  Price  quoted  $9000. 

6.  Vertical  axial  flow  turbine  with  a  capacity  of  125  cu.  sec.  ft.  against  a 
head  of  6  ft.,  with  69  in.  runner,  speed  65  r.p.m.  direct  geared  to  a  150  h.p. 
485  r.p.m.  vertical  motor  and  having  a  guaranteed  efficiency  of  60%. 
Price  quoted  $6100. 

7.  Same  as  (6)  but  for  75  cu.  sec.  ft.  of  capacity  and  57  in.  runner  and 
100  h.p.  motor.     Price  quoted  $4900. 

8.  One  40  in.  horizontal  centrifugal  pump  having  a  capacity  of  125  cu. 
sec.  ft.  against  a  total  head  of  u  ft.,  belted  to  a  250  h.p.  motor.     Pump 
efficiency  guaranteed  50%.     Price  quoted  $6100. 

9.  Same  as  (8)  but  for  6  ft.  head,  50%  efficiency,  and  belted  to  a  150 
h.p.  motor.     Price  quoted  $3400. 

70.  Operating  Costs  of  Centrifugal  Pumps.  —  The  material 
consumed  in  the  case  of  pumps  is  power  together  with  a  very 
little  oil  and  waste.    The  amount  of  power  used  depends  upon 
the  pump  efficiency  and  the  cost  of  the  power  or  its  price.    The 
efficiency  of  pumps  varies  both  with  the  size  of  the  pump  and 
the  head   pumped  against,  larger  pumps  and  greater  heads 
giving  higher  efficiency,  according  to  the  following  table. 


OPERATING   COSTS   OF    CENTRIFUGAL   PUMPS       89 


TABLE  32 

EFFICIENCIES 

CENTRIFUGAL  PUMPS 


HEAD  IN  FEET 

Capacity, 

g.p.m. 

10' 

12' 

is' 

20' 

25' 

30' 

40'  to  70' 

50 

22 

23 

24 

25 

29 

30 

33 

75 

25 

26 

27 

29 

33 

35 

36 

IOO 

28 

30 

32 

34 

36 

38 

40 

150 

32 

34 

35 

37 

40 

42 

45 

200 

36 

37 

39 

42 

45 

48 

50 

300 

40 

4i 

43 

45 

47 

51 

53 

400 

42 

44 

46 

48 

50 

54 

56 

500 

44 

46 

48 

50 

53 

56 

58 

600 

46 

48 

50 

52 

54 

58 

60 

700 

48 

5° 

52 

54 

57 

60 

62 

800 

5° 

52 

54 

56 

61 

63 

64 

900 

52 

54 

56 

59 

62 

63 

64 

1,000 

53 

55 

57 

59 

62 

64 

65 

1,200 

54 

56 

58 

60 

64 

65 

66 

I,4OO 

55 

57 

59 

62 

65 

67 

68 

1,  60O 

57 

59 

61 

63 

65 

67 

68 

PUMP  EFFICIENCY  CONTINUED 

s' 

10'       15' 

25' 

35' 

So' 

6,OOO 

50 

6^       65 

70 

70 

70 

7,OOO 

50 

61       65 

70 

70 

70 

8,000 

50 

62       66 

70 

70 

70 

10,000 

50 

63       67 

70 

73 

15,000 

50 

63.5      68 

70 

74 

20,000 

50 

64       69 

70 

75 

30,000  to 

60,000 

50 

65       70 

70 

BASIC  COSTS 


2000   4000   6000   8000   10000  12000  14000  16000  18000  20000 
G.  P.  M. 

FIG.  8.  —  Efficiencies  of  centrifugal  pumps. 
At  10  ft.  head,  the  variation  of  pump  efficiency  with  capacity 

y  =  67  -  -^=, (49) 


is 


where    y  =  efficiency  in  per  cent, 

and      M  =  the  pump  capacity  in  thousands  of  gallons  per 

minute  capacity. 

With  increased  head  the  pump  efficiency  increases  by  the 
following  per  cents: 

Head  .........................     10    12     15     20    25        30    40    50 

Per  cent  increase  in  efficiency.  ...       o      4      9     15     18.5     21     25     28 

Corresponding  to  the  equation 

160  ,    x 

......  (50) 


where    yp  =  per  cent  increase  in  efficiency, 
and          h  =  the  head  in  feet. 

The  lubricating  oil  used  amounts  to  about  o.ooi  pint  per 
water  horse  power  hour. 


FRACTIONAL  LOAD   EFFICIENCY  91 

71.  It  is  often  assumed  that  the  cost  of  attendance  of  cen- 
trifugal pumps  is  zero.  This  is  hardly  correct,  although  this 
cost  for  centrifugal  pumps  is  very  small.  It  depends  upon  the 
general  design  of  the  installation  normally  being  about 


<*' 


where    A  =  cost  of  attendance  in  cents  per  h.p.h.  as  of  1910, 
and        M  =  the  h.p.  required  to  drive  the  pump. 

It  must  be  borne  in  mind  that  all  costs  are  as  of  1906-10. 
To  get  the  attendance  cost  as  of  today  divide  by  the  relative 
value  of  gold  now  as  compared  with  the  date  above,  as  indi- 
cated by  the  relative  purchasing  power.  Thus  if  the  purchasing 
power  of  gold  now  is  only  half  of  what  it  was  then,  all  costs 
now  would  be  double  that  given  above. 

Based  on  the  study  of  the  efficiency  of  pumps  at  fractional 
loads  covering  the  tests  of  several  hundred  pumps  varying  in 
size  from  i  inch  to  12  inches  and  consisting  of  four  different 
makes  and  some  twelve  different  types,  we  have  obtained  the 
following  average  values  of  the  per  cent  of  full  load  efficiency 
at  fractional  loads  viz.  : 

Per  cent  of  full  load  ............  o      25      50      75       100      125     150 

Per  cent  of  full  load  efficiency.  ..  o    43.6    75     93.7     100     93.7     75 

Corresponding  to  the  equation 

yp  =  F(2  -  o.oiF)      ......  (52) 

where    yp  =  per  cent  of  full  load  efficiency, 
and         F  =  per  cent  of  full  load  efficiency, 


92  BASIC   COSTS 

72.   Motors  and  Generators. 

TABLE  33 
INDUCTION  MOTORS  WITH  STARTERS 

PRICES,  1912 
2  and  3  phase  60  cycles  110-220  volts,  squirrel  cage  type 


SIZE, 

R.P.M. 

H.P. 

1800 

I2OO 

900 

720 

600 

i 

$33-oo 

$40  .  oo 

$47  •  oo 

i 

40.00 

50.00 

63.00 

$77-00 

2 

50.00 

58.00 

75-oo 

91  .00 

$110.00 

3 

57-oo 

71  .00 

94.00 

115.00 

140.00 

5 

70.00 

90.00 

117.00 

143.00 

170.00 

1\ 

i35-oo 

160.00 

195.00 

220.00 

267.00 

10 

160.00 

195.00 

230.00 

265  .  oo 

300.00 

15 

192.00 

225.00 

270.00 

306  .  oo 

348.00 

20 

235-00 

272.00 

305.00 

330.00 

390.00 

25 

255-00 

305.00 

340.00 

377-oo 

430  .  oo 

30 

305-00 

340.00 

375-oo 

420  .  oo 

465  .  oo 

40 

350.00 

390.00 

435-00 

481  .00 

535-oo 

50 

400  .  oo 

440.00 

485.00 

535-oo 

595-oo 

75 

515-00 

565.00 

610.00 

664  .  oo 

720.00 

100 



680  .  oo 

724.00 

780  .  oo 

832.00 

ISO 

960.00 

IOIO.OO 

1070.00 

0 

$1200 
1000 
800 
600 
400 
200 


20  40          60          8< 

FIG.  9.  —  Prices  of  squirrel  cage  motors. 


100         120         140         160         180       200 
H.  P. 


73. 


MOTORS 

TABLE  34 
INDUCTION  MOTORS  WITH  STARTERS 

PRICES,  1910-12 
Slip-ring  type  1 10-550  volts,  3  phase  60  cycle 


93 


SIZE, 

R.P.M. 

. 

H.P. 

1800 

I2OO 

goo 

720 

600 

I 

$60  oo 

$77  .00 

I 

80.00 

104.00 

$113.00 

2 

IO4.   OO 

113  oo 

136  oo 

122   OO 

138  oo 

164  oo 

5" 

I?1?    OO 

175  OO 

188.00 

7! 

210  do 

224  oo 

250.00 

IO 

22C   OO 

260  oo 

280  oo 

$300  oo 

15 

20 

275.00 

300.00 
34.0  oo 

330.00 

380  oo 

360.00 

410.00 

$380.00 

440.00 

2C 

380  oo 

425.00 

468.00 

405  .OO 

30 

AQ 

425.00 

s;O5   OO 

475-00 
562  oo 

515-00 

608  oo 

550.00 

640  oo 

r<D 

575.OO 

640  oo 

680.00 

710  oo 

75 

725.00 

790  .  oo 

830.00 

860.00 

IOO 

920  oo 

960  oo 

QQO   OO 

As  previously  noted,  these  prices  are  as  of  1910-12. 
obtain,  for  example,  prices  as  of  1917,  divide  by  0.65. 


To 


94 


BASIC   COSTS 


74. 


TABLE  35 
DIRECT  CURRENT  MOTORS 

PRICES,  IQIO-I2 
Open  shunt  wound  type  115-230  volts 


H.p. 

Speed 

Price 

I 

1500 

$65.00 

i-5 

1400 

75.00 

2-5 

1400 

IOO.OO 

3-5 

1700 

100.00 

3-5 

1300 

128.00 

5 

1550 

128.00 

5 

I  TOO 

160.00 

7-5 

1500 

160.00 

7-5 

1000 

200.00 

10 

1400 

205  .  oo 

10 

IIOO 

233.00 

IS 

1400 

245  .  oo 

15 

IOOO 

284.00 

20 

1350 

284.00 

20 

900 

346.00 

25 

I2OO 

346.00 

25 

800 

440  .  oo 

30 

IOOO 

440.00 

30 

700 

500.00 

35 

950 

516.00 

35 

700 

570.00 

40 

950 

500.00 

40 

600 

665  .  oo 

50 

IOOO 

600.00 

50 

600 

785.00 

NOTE. — The  price  of  compound  wound  motors  is  5%  higher. 


75. 


MOTORS  AND   GENERATORS 


95 


Cost  of  Motors  <2  V.  ' 

& 

X 

& 

x 

r^ 

/ 

^ 

^ 

X 

x 

X 

^ 

S 

/ 

'x 

<s 

/ 

/ 

/ 

/, 

'/ 

fr 

10    20    30    40     50     60    70    80    90   100 
-  H.  P. 

FIG.  10.  —  Prices  of  direct  current  motors. 


$30 


25 


20 


10 


\ 

\ 

\ 

\ 

\ 

^ 

"**«-. 

—  , 

-—  ». 

—  ~ 

^   m 

-~— 

—  —  ^_ 

-~™™ 

—  ~— 

= 

100    200   300    400    500    600    700    800    900   100* 
K.  V.  A. 

FIG.  ii.  —  Prices  of  A.  C.  generators. 

TABLE  36 
A.  C.  GENERATORS 

PRICES,    IQIO-I2 

60  cycle  three-phase  220-550  volts.     Direct-connected  type 

Size,  kv.a 10         25        50        75       100      150      250      500       750     1000 

Price  per  kv.a.  .  .$25.00   20.00   16.50   14.75    14.00   12.85    11.50   10.00    9.60      9.40 


96 


BASIC  COSTS 


^  74, 


(53) 


This  corresponds  almost  exactly  to  the  equation 
C  =  -^- 

VM 

where    C  =  cost  in  dollars  per  kv.a., 
and       M  =  the  size  in  kv.a. 

TABLE  37 
TRANSFORMERS 
PRICES,  1910-12 

IIOO-22OO  to  IIO-22O  Volts 

Size,  kw. ...0.6          i           1.5  2  2.5           3 

Price $26.00    32.00    39.00  46.00  51.00     57.00 

Size,  kw..  .  .      10               15  20              25              30 

Price $133.00      184.00  229.00      269.00      314.00 

76.  Operating  Costs  of  Electric  Machinery.  —  The  efficiency 
at  full  lead  of  three  phase,  squirrel  cage  motors,  no  to  550 
volts,  is  independent  of  the  speed  but  varies  with  the  size  as 
given  below. 


4 
68.00 


5 
80.00 

40 
395-00 


7-5 
108.00 

50 
474-oo 


Cost  of  Transformers  ^ 

^^ 

^ 

. 

^ 

^ 

^ 

^ 

_x 

^ 

^ 

^ 

^ 

^ 

^ 

xx 

x^ 

X0 

xx*"<^ 

X 

x 

^ 

5     10    15    20    25    30    35    40    45    50 
Kw. 

Size  in  h.p 

Efficiency  in  per  cent. 
Size  in  h.p 


FIG.  12.  —  Prices  of  transformers. 

TABLE  38 
EFFICIENCIES 

Induction  motors  3  phase  60  cycle 
.12  3          5        7.5     10       15     20      25 

.78.4  82.5     84.2     85.8     87.2  87.9  88.6  89.1  89.4 
•  3°      35        40        50       60      75     100    150     200      250 


Efficiency  in  per  cent..  .89.7  89.85  90.05  90.25  90.4  90.6  90.8  91.1  91.24   91.3.3 


OPERATING  COSTS  OF  ELECTRIC  MACHINERY     97 


IbU 
140 
130 
120 
110 
100 

Joo 

o 
S80 

I™ 

KGO 

50 
40 
30 
20 
10 
0 

1 

1 

I 

1 

Efl 

'.  =  ( 

2.2- 

gr 

>. 

f 

'/        ' 

t 

>/x 

\    '* 

£ 

X  ; 

I 

X  •' 

|/ 

• 

/ 

• 

.*_ 

= 

=E 

—  — 

^± 

X 

*—  — 

—~~ 

~f* 

t***^ 

72    73    74    7.5    76    77    78    79    80   81    82    83    84    85    86    87 
Per  cent.  Efficiency 

FIG.  13.  —  Efficiencies  of  A-c.  60  cycle  motors. 

This  corresponds  to  the  equation 

1^.8 
y  =  92.2  — 


89    90   91 


(54) 


where    y  =  the  per  cent  efficiency, 
and      M  =  the  size  in  h.p. 

It  should  be  borne  in  mind  that  such  data  as  we  present  are 
based  on  the  study  of  the  results  of  tests  of  our  very  best 
and  largest  manufacturers.  The  data  themselves,  together 
with  the  curve  of  values  adopted,  is  shown  in  Fig.  13.  The 
discrepancies  in  the  data  are  due  to  redesign  of  some  units  be- 
fore the  like  service  could  be  given  to  others;  and  to  some 
error  in  test,  which,  though  small,  is  nevertheless  quite  signifi- 
cant and  finally  to  commercial  exigencies  such  as  the  use  of  a 
frame  designed  for  one  given  size  and  speed,  for  a  different 
size  and  speed.  Such  adaptations  of  one  frame  to  several  sizes 
generally  give  greater  economy  of  production  but  at  a  re- 
duction in  worth  of  the  unit  due  to  reduced  efficiency  and 
power  factor. 


98 


BASIC  COSTS 


The  efficiency  of  generators  at  full  load  for  3  phase  alter- 
nating current  of  60  cycles  at  from  200  to  550  volts  varies 
with  size  as  follows: 


TABLE  39 

EFFICIENCIES  OF  A-c.  GENERATORS 
60  cycles  220  to  550  volts 

Size,  kv.a 50        75        100      125  150      200      300      500 

Per  cent  efficiency 88.5    90.0    90.9    91.5  91.9    92.6    93.2    93.9 

Size,  kv.a 750      1000     1250     1500  2000     3000    4000     5000 

Per  cent  efficiency 94.4    94.7    94.9    95.0  95.2    95.4    95.5    95.6 


These  values  correspond  to  the  equation 


y  =  96.4  - 


55 


(55) 


77.   At  fraction  loads  the  per  cent  of  full  load  efficiency 
obtained  from  motors  and  generators  is  as  follows : 


100 

99 
98 
97 
96 


or 
£3  95 


93 

§92 


500       1000       1500       2000       2500       3000      3500       4000      4500    5000 
K.  V.  A. 

FIG.  14.  —  Efficiencies  of  a-c.  generators. 


INTERNAL  COMBUSTION  MOTORS  99 

TABLE   40 

PER  CENT  OF  FULL  LOAD  EFFICIENCY  OF  INDUCTION  MOTORS 
3  phase  60  cycle  210-550  volts 

Per  cent  of  full  load  ..............     125  100  75  50  25 

Per  cent  of  full  load  efficiency  ......       97  100  100  97  90 


These  values  correspond  to  the  equation 


I00  - 


.   (56) 


4OO 

where    yp  =  per  cent  of  full  load  efficiency, 
and         F  =  per  cent  of  full  load  efficiency. 

TABLE  41 
PER  CENT  OF  FULL  LOAD  EFFICIENCY  OF  A-c.  GENERATORS 

3  phase  60  cycle  220-550  volts 
Per  cent  of  full 

load  .......      125      100      90        80        70        60        50        40        30        25 

Per  cent  of  full 

load  efficiency  98  .  5     100    99.5     98.5     97.0    95.0    92.0    89.0    85.0    82.5 

This  corresponds  to  the  equation 

(100  -  FY  ,    x 

yp  =  100  -  ^  -  '—       ....    (57) 
320 

The  attendance  cost  of  motors  and  generators  of  1910-12  is 
approximately 

A  =  °>  ........   (58) 


78.  Internal  Combustion  Motors.  —  The  prices  of  gas 
engines  depends  on  whether  they  are  two  or  four  cycle,  Otto 
cycle,  Diesel,  or  semi-Diesel,  as  well  as  on  the  design  and 
workmanship.  Some  engines  are  needlessly  elaborate,  others 
so  unsubstantially  built  that  they  are  made  to  sell  rather  than 
to  use.  The  two  stroke  engine,  except  for  the  Diesel  or  semi- 
Diesel  cycle,  is  not  of  practical  value,  and  no  data  thereon  is 
therefore  given. 

The  sizes  and  corresponding  prices  (1912)  for  one  make  of 
well-designed  and  well-built  type  of  gas  engines  is  given  below: 


IOO 


BASIC  COSTS 


TABLE  42 
STATIONARY  OIL  ENGINES 

PRICES,  1910-12 
Heavy  Duty  —  Otto  cycle 


SIZE 

HORIZONTAL 

VERTICAL 

No.  cy. 

Total 

Per  h.p. 

No.  cy. 

Total 

Per  h.p. 

2 

$180.00 
2  7O   OO 

$9O.OO 
90  oo 

I 

$l8o.OO 

$90  .  oo 

4 
6 

8 

12 

18 

360.00 
500.00 

630  .  oo 
810.00 
i  080  oo 

90.00 

83.33 
78.75 
67.50 

60  oo 

I 

I 

I 

270.00 
360.00 

72O.OO 

67  .  50 
6o.OO 

60.00 

25 

OC 

1350.00 

54.00 

I 
2 

1080.00 
I3C.O.OO 

43-25 
^4.00 

•22 

i 

1660  oo 

<2  .OO 

36 

/in 

2 
I 

2160.00 
1980  oo 

6O.OO 
4.O    ^O 

3 

1980  .  oo 

55-oo 

So 

i  or  2 

2520.00 

50.40 

2 
I 

2160.00 
2520.00 

43-20 
50.40 

64 

2 

•j  2AO   OO 

CO    CO 

7e 

3 

3260.00 

43.50 

7t? 

I 

3870  oo 

51  .60 

80 
IOO 

2 
2 

3870.00 

5000  .  oo 

48.38 
C.O.OO 

i  or  2 

5000  .  oo 

50.00 

IOO 

4 

4320.00 

43-  20 

Price  of  Engines  .^ 

^ 

x^" 

s* 

^^ 

^^ 

x^ 

** 

x^ 

S^~ 

^ 

x^ 

Me 
E 

mH 
iffine 

rizo 
Cost 

ital- 

s   X 

>^ 

>xx 

x^ 

,X 

^ 

<^ 

S^ 

xMean  Vertical 
'  —  Engine  Costs 

-X"* 

^ 

x* 

^ 

j^r 

x^ 

^ 

^ 

^ 

^ 

^ 

^ 

X* 

xx 

^ 

* 

^ 

^ 

)           10          20 

30          40           50          60           70          80           90         10 
H.P. 

FIG.  15.  —  Prices  of  oil  engines. 


INTERNAL  COMOTSTlON '< 


101 


Fernold  and  Orrok  in  their  "  Engineering  of  Power  Plants  " 
give  the  cost  of  producer  gas  engines  as  follows : 

TABLE   43 

PRODUCER  GAS  ENGINES 
PRICES,  1908-10 


PRICE  F.O.B.  FACTORY 


dL£.£.,   n.f. 

Total 

Per  h.p. 

20 

$1,000.00 

$50.00 

60 

2,800.00 

46.67 

75 

3,6lO.OO 

48  .  10 

3,400.00 

42.50 

80 

3,250.00 

40.70 

3,830.00 

47.90 

8s 

4,150.00 

48.90 

°o 

3,500.00 

41.80 

IOO 

4,925.00 

49-25 

(4,950.00 

45-oo 

no 

{  4,960.00 

45  -!0 

1^  4,200.00 

37-5° 

130 

5,250.00 

40.40 

135 

6,600.00 

48.80 

160 

/  5,500.00 

\  6,100.00 

35-oo 
38.10 

250 

6,650.00 

26.60 

400 

(12,000.00 
12,800.00 

30.00 
32.00 

600 

17,400.00 

29.00 

IOOO 

33,750.00 

33-75 

2OOO 

64,850  .  oo 

32-43 

79.   The  prices  of  4  cycle  Diesel  engines  run  as  follows ; 

TABLE   44 

DIESEL  ENGINES,  FOUR  CYCLE 
1910-12 


Rated 

Price  of  engine 

Price  per  h.p. 

60 

$4,620.00 

$77-oo 

90 

6,060  .  oo 

67.00 

no 

7,000.00 

63  50 

150 

9,000.00 

60.00 

200 

11,500.00 

57-50 

250 

14,000.00 

56.00 

300 

16,500.00 

55-oo 

400 

21,200.00 

52.80 

500 

25,800.00 

51  .60 

750 

37,200.00 

49.60 

IOOO 

48,300.00 

48.30 

102 


SVr  ! 


BASIC  COSTS 


9YU,UUW 

60,0000 

3  50,0000 
24<M)000 

I 

*  30,0000 
20,0000 

X" 

X 

x 

x^ 

^ 

X 

X 

p 

x" 

X 

X 

x 

x 

X 

x 

x* 

X 

X 

x^ 

.10,0000 
n 

x 

X 

x 

<xp^ 

200    400    600    800   1000   1200   1400   1600   1800   2000 
H.  P. 

FIG.  1 6.  —  Prices  of  producer  gas  engines. 

The  prices  of  large  two-cycle  Diesel  engines  is  reported  as 
low  as  $35  per  h.p. 


$ou,uuu 

40,000 

•j 

1 

£30,000 

o 

1 

^20,000 

-X 

x^ 

x^ 

x^ 

x^ 

^^ 

^* 

•^ 

x^ 

x**^ 

x^ 

x^^ 

x^ 

x^* 

^f 

^ 

p^ 

1U,UUU 
0 

^x 

xx 

100         200        300        400         500         600        700        800         900     1000 
H.P. 

FIG.  17.  —  Prices  of  4-cycle  Diesel  engines 


INTERNAL  COMBUSTION  MOTORS 


103 


80. 


Of^O          w          "^         ON 

fOOO    to          rj-          r^        vo 


cs    o    ^ 

PO  O    to 


IO    <N     IO 


totot^cs  ty>          O          OO 

<N-^-cot^vO  O  O 


M  <N  00 


t>.lOOooO  OO 


tOMOO  00 


<N  to  O 

MHO 


IO<N«NIN          ON 
r^tOMOO          M 

t^.  CO  H 


toOtoio         to         M          ^ 

o»vOr^-ON         PO        oo          to 
OO  fO  M  M  cs 


OQtOCN  M  O  ^  "4" 

O^    t^-    t^»    M  to  O  *"*  ON 


8|  S| 

E      H      H      U 


.1 


M 
ss 

to 

1 


J  SH 

If  «•* 

<U     -     '• 


as  w 
tfal 


11 


.   5^ 


BS.S.| 

8-88.S. 


.2  ^  ^ 
^  ^  oo 
0  "o  Q 


H**5« 

=  53g.«13 


S  8  8  S3  g 

cj    ^J    u    U  *S 

SSS2| 

I  <u 


104  BASIC  COSTS 

In  Guldner's  "Internal  Combustion  Motors"  (D.  Van 
Nostrand  Co.)  are  given  very  extensive  data  on  gas  engines 
from  which  the  data  given  in  Table  45  on  illuminating  and 
suction  gas  producer  installations  are  quoted. 

81.  The  variation  of  thermal  efficiency  of  gas  and  oil  engines 
with  size,  of  good  modern  design  is  as  follows : 

TABLE  46 
THERMAL  EFFICIENCY 

GAS  AND  OIL  ENGINES 

H.p.  per  cylinder. ..   10        20       30       40        50       60       80        100      125      150 
Full  load  efficiency, 
percent 20.8    21.8    22.5    22.9    23.2    23.5    24.0    24.3    24.6    24.9 

H.p.  per  cylinder 200        300        400        500        600        800        1000 

Full  load  efficiency,  per  cent.  25.3       26.0      26.4       26.7       27.0       27.5       27.8 

These  values  correspond  to  the  equation: 

y  =  i7-3  +  3-5  log  M, (59) 

where  y  =  per  cent  efficiency, 
and     M  =  the  h.p.  per  cylinder. 

The  efficiency  of  a  two  cylinder  engine  is  no  better  than  a 
single  cylinder  engine  of  half  its  size,  though  its  regulation  is 
usually  better. 

82.  At  fractional  loads,  the  following  per  cents  of  full  load 
efficiency  are  obtained: 


TABLE  47 

PER  CENT  OF  FULL  LOAD  EFFICIENCY  or  GAS  AND  OIL  ENGINES  AT 
FRACTIONAL  LOADS 

"Per  cent  of  full  load 100    90    80     70    60     50    40    30     20     10    o 

Per  cent  of  full  load  efficiency.  ..    100    99    96    91     84     75     64     51     36     19    o 

These  values  correspond  to  the  equation 

yp  =  F(2  -  F), (60) 

where  yp  =  per  cent  of  full  load  efficiency, 

and       F  =  per  cent  of  full  load  expressed  as  a  fraction. 


INTERNAL  COMBUSTION  MOTORS 
The  attendance  cost  (A)  per  bo.h.p.h.  is 

A  -  °'0°5, 

•*X      ~  "  7 

VM 


105 


(61) 


where  (A)  is  expressed  as  usual  in  dollars. 

The  amount  of  lubricating  oil  used  is  usually  0.005  to  0.007 
pints  per  br.h.p.h. 


32 

r>n 

^ent,  Efficiency 
2  £  g 

Ml—1 

.  

.  

—  —  - 

.,      — 

SS~* 

—  —  - 

• 

—  —  • 

—     — 

* 

•• 

.•       " 

^ 

r 

^ 

^ 

,^--1 

91 

X 

/ 

/ 

7 

on 

"U0          100        200        300        400         500        600         700        800         900       100 

H.r. 

FIG.  1 8.  —  Full  load  efficiency  of  gas  and  oil  engines. 

The  maintenance  of  gas  and  oil  engines  is  usually  expressed 
as  between  2  %  and  3  %  of  the  first  cost  of  the  engine.  This  is 
not  good  practice  because  the  first  cost  of  a  certain  engine  in 
use  is  a  constant,  while  the  cost  of  maintenance  varies  with 
the  price  of  wages,  or,  if  you  please,  with  the  value  of  the  dollar. 
This  method  could  be  used  provided  that  the  first  cost  is  first 
multiplied  by  the  relative  value  of  the  dollar  (indexed)  before 
the  2  %  or  3  %  is  taken.  Otherwise,  it  would  usually  be  too 
low,  the  error  greatly  increasing  with  time. 

83.  The  variation  of  full  load  efficiency  of  Diesel  engines 
with  size  is  as  follows: 


io6  BASIC  COSTS 

TABLE  47 
FULL  LOAD  EFFICIENCY  OF  DIESEL  ENGINES 

H.p.  per  cylinder  .......  10  15      20        25       30      35       40       50      60       70 

Full  load  efficiency,  per 

cent  ................  24.0  26.3   27.7    28.5    29.1    29.5    29.8   30.2    30.5    30.7 

H.p.  per  cylinder  .......  80  100     125      150    200 

Full  load  efficiency,  per 

cent  ................  30.9  31.1  31.2  31.3  31.4 

These  correspond  to  the  equation 


i  , 

=  —  5-tan-M  —  ).    ......    (62) 


20          40          60          80         100        120         140        160         180       200 
H.P. 

FIG.  19.  —  Full  load  efficiency  of  Diesel  engines. 

At  fraction  loads,  the  per  cent  of  full  load  efficiency  obtained 
is: 

TABLE  49 

PER  CENT  OF  FULL  LOAD  EFFICIENCY  OF  DIESEL  ENGINES  AT  FRACTIONAL 

LOADS 

Per  cent  of  full  load .  .100    90    80    70    60    50    40    30     20    10 

Per  cent  of  full  load 100    99    97    93    88    82     74    65     54    42 

The  equation  for  these  values  is 

y  =  100  —  72(100  —  F)2 (63) 

All  the  above  data  is  based  on  the  guarantees  of  manufac- 
turers together  with  a  careful  study  of  the  most  reliable  tests 
reported.  The  accuracy  should  be  better  than  can  be  obtained 
from  any  single  test. 


STANDARD   PIPE 


107 


84.   We  give  below  prices  of  various  sized  pipe  as  of  1912, 
inasmuch  as  we  shall  have  use  of  such  data  hereafter. 

TABLE   50 
STANDARD  PIPE 

PRICE   (1912)   PER   100  FEET,  RANDOM   LENGTHS 


S^ZE, 

INCHES 

STANDARD 

EXTRA  STRONG 

DOUBLE  EXTRA  STRONG 

FLANGE 

Black 

Galv. 

P.E. 

C.&T. 

Plain  ends 

Threads 

1 

$5  25 

$c,    en 

i 

T    oC 

$2    7C, 

5  25 

5  •  5° 

2    7C 

c  .  co 

i 

2    7  C. 

7     85 

c.  25 

5.50 

$18.00  * 

1 

3-20 

4-45 

5-50 

5-8o 

2O.OO 



I 

4-55 

6.36 

8.10 

8.50 

25.00 

$0.06 



jl 

6    2C, 

8.70 

ii  .00 

ii  .60 

35-oo 

if 

7-5° 

10.40 

13.20 

13.90 

45-oo 

0.08 



2 

9-55 

I3-50 

17-75 

18.65 

60.00 

O.IO 

$0.90 

2? 

15-75 

22.  2O 

29.15 

30.60 

90.00 

0.15 

1.20 

3 

20.70 

29.10 

37-75 

39-65 

125.00 

O.2O 

1.35 

3^ 

28.15 

38.75 

51.40 

54-oo 

165.00 

0.25 

1.50 

4 

32.00 

44-05 

58.00 

60.90 

190.00 

o-35 

I.  80 

4l 

40.00 

54-50 

99.40 

0-55 



5 

44.60 

60.80 

IIO.OO 

260.00 

0-55 

2.  2O 

6 

57-8o 

78.80 

148.00 

360.00 

0.70 

2.40 

7 

80.00 

153-00 

240.00 

0.85 

3-45 

8 

85  25 

163  .OO 

271.  oo 

I    OO 

300 

116.00 

221  .OO 

4QCJ 

10 

i  20.00 

227.OO 

394  .  oo 

I  .  C.O 

5C.  C. 

ii 

292  .OO 

12 

i53-oo 

292.OO 

466.00 

2.50 

7.50 

io8 


BASIC  COSTS 


TABLE  51 

STANDARD  CASING 

PRICE  (1912)  PER  100  FEET,  RANDOM  LENGTHS 


O.D. 

I.D. 

Wt.  per  ft. 

Casing 

C-it 

Thread 

Flange 

2|* 

ti- 

2.82 

$12.90 

O.IO 

0-15 

1-35 

3i 

ll 

3-45 

15-57 

O.IO 

O.2O 

1.42 

3* 

4-45 

18.72 

0.15 

0.25 

1.  80 

4 

3l 

5-56 

22.68 

O.2O 

o-35 

2.16 

4! 

4* 

6.36 

25-25 

0.25 

o-55 

2.30 

5 

4f 

7.80 

31-59 

0.30 

0.60 

2.50 

6 

Si 

10.46 

37-00 

0-35 

0.70 

3-io 

7 

6| 

12.34 

47-54 

0-45 

0.85 

4.60 

8 

7f 

I5-4I 

58.25 

0.50 

i  .00 

5-25 

10 

9l 

21  .90 

84.75 

0.75 

i-5o 

6.90 

12 

n| 

30-35 

131.00 

1-25 

2.50 

9.00 

$200 
& 

LJ 

3  160 


>!-St 


Pipe 


120 


1 


Utd  Cas 


i. 


Wood 


6  8  10          12  U  16  18          20 

Dia.  of  Pipe,  Inches 

FIG.  20.  —  Prices  of  pipe. 


WOOD   PIPE 


log 


TABLE  52 

WOOD  PIPE 

(1912) 


PI 

UCE  PER 

100  FEET 

FOR  HEA 

DS  OF 

SIZE 

So' 

100' 

ISO' 

200' 

2So' 

300' 

350' 

4oo' 

2" 

$9-75 

$10.00 

$10.28 

$11.11 

$11-95 

$12.75 

$14-45 

$16.40 

3 

ii-QS 

12.50 

12-75 

13-35 

13.90 

14-75 

17.22 

18.90 

4 

14-45 

15.00 

15.28 

18.60 

J9-45 

21.  II 

23-33 

25-55 

6 

i8.33 

21  .40 

23-33 

25.28 

27.50 

30.00 

32-33 

33-33 

8 

22.50 

27.78 

3I-40 

35-55 

38.60 

41.40 

44-45 

51-67 

10 

29.72 

35-83 

40.83 

46.95 

50.85 

55-83 

59-17 

68.33 

12 

33  -61 

43.06 

50-55 

56.11 

63.06 

7O.OO 

81.40 

90.00 

14 

44.00 

6O.OO 

T6 

56.00 

67.00 

18 

72.OO 

20 

87.00 

24 

83.00 

IO2  .OO 

no 


BASIC   COSTS 


TABLE   53 
W.  I.  PIPE  FRICTION 

FEET  PER   100  FEET  OF  PIPE 


G.P.M. 

SIZE  OF  PIPE  IN  INCHES  OF  DIAMETER 

i 

U 

i 

2* 

3 

4 

5 

6 

8 

IO 

12 

5 

10 

15 

20 
25 
30 
40 

SO 
70 
100 
120 
ISO 
175 
200 
250 
300 

350 
400 
500 

750 
1000 

1250 
1500 

2OOO 
2500 
3000 
3500 
4000 
4500 
5000 

2.32 
8.40 
18.9 
30.1 

45-5 
64.0 
109.0 

0.28 

1.02 
2.25 
3-70 
5-60 

7.8 

13-3 
20.2 

37-6 
73-0 

0.08 
0.36 
0.81 
1.29 
i  .96 

2-73 
4.68 
7.10 
13.2 
25-6 
36.0 
54-0 

0.06 
0.12 
0.25 

0-43 
0.66 
0.92 
1-57 
2.38 
4.42 
8.6 

I2.O 
18.7 
23-7 
30-9 

O.II 

0.18 
0.27 
0.38 
0.65 
0.98 
1.83 
3-52 
4-97 
7.72 

9-75 

12.8 

19.7 
27.1 

0.16 

0.24 

o-45 
0.88 

I  .  22 
1.82 
2.4O 
3.12 
4.80 
6.70 
8.80 

U-3 

17.2 

0.15 
0.29 
0.41 
0.63 
0.84 
1.  06 
I.  60 
2-25 
2.99 
3-8l 

5-8 
12.3 

::::: 

0.18 
0.23 

0-34 
0.44 
0.66 
0.92 

I.  21 

1.58 

2-33 
4.87 
10.3 
13.0 

o.  16 
o.  26 
0.29 
0.40 
0.58 

I  .  21 

2-51 
3-18 
4.48 
7.65 

1 

0.19 

o-39 
0.83 
1.03 
1.49 
2.50 
3-8i 
5-3° 
7.20 

0.17 

o-34 
0-43 
0.61 

1.02 
I.S6 
2.42 
2.80 
3-80 
4.82 
5.82 

• 

Harding  &  Willard:  *  When  slightly  rough  add  15%. 

When  very  rough  add  30%. 
»,  *  John  Wiley  &  Sons,  Inc. 


WOOD   AND   RIVETED   STEEL  PIPE 


in 


For  large  sized  pipe  the  following  data  are  given  at  Chicago, 
including  the  cost  of  laying  but  less  haulage. 

TABLE  54 

WOOD  STAVE  PIPE 

COST  (1912)  PER  FOOT 


HEAD  IN  FEET 

SIZE, 

INCHES,  DIAM. 

25 

SO 

TOO 

200 

12 

$0.42 

$0.49 

$0.63 

$0.85 

18 

0.69 

0.8o 

1.02 

I  .46 

24 

0.79 

0.91 

I.I4 

1.61 

30 

0.96 

I.  12 

1.44 

2.06 

36 

1.19 

1.40 

1.82 

2.65 

42 

1.40 

1.68 

2.23 

3-33 

48 

i-S5 

1.85 

2.46 

3.67 

54 

2.23 

2.62 

3-43 

5-02 

60 

2.85 

3-35 

4-37 

6.40 

66 

3-21 

3-8i 

5-oo 

7.38 

72 

3-65 

4-38 

5-83 

8-73 

Wood  pipe  has  made  good  in  practice  and  has  come  into 
very  general  use.  The  friction  of  unplaned  wood  pipe  is  about 
equal  to  riveted  steel  pipe,  but  that  of  planed  wood  pipe  is 
given  at  only  60%  to  80%  of  that  of  riveted  steel  pipe. 

TABLE   55 
RIVETED  STEEL  PIPE 
COST  (1912)  PER  FOOT 


Size, 


GAGE  THICKNESS 


inches,  diam. 

14 

12 

IO 

8 

6 

i 

A 

§ 

I  2 

$0    ^2 

$0  18 

$0.44 

18 

24 

0-57 

0.65 

o.St; 

$0.78 
1  .04 

$0.98 
1.28 

$1-55 

$i  .99 

•?o 

I    27 

I    $Q 

I    Q3 

2    46 

$?  04 

l6 

I  .  CC 

I   03 

2  .  ^O 

2    O2 

3  58 

4.2 

I.6l 

2.18 

2.66 

2  .  27 

4.12 

48 

2.48 

3.03 

3.83 

4.66 

C4 

2.80 

3.41 

4-  20 

5-21 

60 

3    70 

4   7^ 

c  .  74 

66 

4-  3^ 

<  .  21 

6.  29 

72 

4-  ^2 

s.66 

6.83 

112 


BASIC  COSTS 


TABLE   56 

TABLE  OF  PIPE  FRICTION 
FOR  LARGE  RIVETED  STEEL  PIPE  IN  FEET  PER  TOO  FEET  OF  PIPE 


Cu   SEC   FT 

SIZE  c 

>F  PIPE 

IN  INC 

HES    OF 

DlAME 

PER 

12 

18 

24 

30 

36 

42 

48 

54 

60 

66 

72 

er 

T    6 

o  •? 

IO 

1  8 

i  .0 

o  3 

O.  I 

1C 

T    8 

O.  i? 

O.  2 

0.06 

2O 

"?•  2 

o  8 

o.  3 

O.  2 

o.o<; 

30 

CO 

1.9 

o-S 

I     A 

0-3 

O   7 

0.17 

O  4 

O.I 

o  16 

o  08 

IOO 

200 

I  .O 

0.56 

0.30 
I  .OO 

O.2 

0.66 

O.  I 

0.47 

0.07 

O    20 

2QO 

o  84 

O    ^O 

The  same  authorities  give  the  approximate  section  of  dams 
not  used  as  weirs,  and  their  approximate  cost  as  follows, 
where  (h)  is  the  height  of  the  dam  in  yards. 


TABLE   57 

APPROXIMATE  COST  OF  DAMS 
1910-12 


Common 

Common 

Type  of  dam 

up  stream 
batter, 

down  stream 
batter, 

Approx.  sec. 
in  sq.  yds. 

Approx.  cost 
per  cu.  yd. 

Approx.  cost 
per  lineal  yd. 

hor.  to  vert. 

hor.  to  vert. 

Earth  .- 

2  to  I 

3  to  I 

2-5A2 

$0.50 

$I.2S# 

Crib  

if  to  if 

\\  to  i 

i  .  5/?2 

i  -5° 

2.25/J2 

Rock  fill  

2  tO  I 

2  tO  I 

2.0/f2 

2.00 

4  .  ooA2 

Masonry,  straight 

or  o  to  i 

vertical 

0.8  to  i 

o.4/f2 

12  .OO 

4  Sohz 

Masonry,  arched. 

0.15  to  i 

0.3  to  i 

15.00 

3-  So/*2 

Tunnels.  —  On  the  average,  the  cost  of  tunnels  including 
timbering  and  lining  is  $15  per  cubic  yard.  The  costs  and 
capacities  of  tunnels  is  given  below. 


CANALS 

TABLE  58 

COST  OF  TUNNELS 

1910-12 


Size  in  ft. 

Sectional 
area, 
sq.  ft. 

Allowable 
velocity, 
ft.  per  sec. 

Capacity, 
sec.  ft. 

Approx. 
slope  per 

1000  ft. 

Approx. 
cost  per 
lih.  ft. 

Cost  per  lin.  ft. 
per  100  sec.  ft. 

4X7 

28 

3-6 

IOO 

0.46 

$16.00 

$16.00 

7X7-25 

5° 

10 

-  500 

2.O 

28.00 

5.6o 

10  X  10 

IOO 

10 

1,000 

1-5 

56.00 

5.6o 

12  X  12.5 

150 

IO 

1,500 

I.I 

85.00 

S-67 

14  X  14-25 

2OO 

IO 

2,OOO 

0.9 

115.00 

5-75 

20  X  25 

500 

IO 

5,000 

0.6 

280.00 

5-6o 

30  X  33 

1000 

10 

10,000 

o-3 

500.00 

5.00 

Canals.  —  In  ordinary  earth  a  velocity  of  2  feet  per  second 
is  commonly  used.     On  this  basis  the  following  data  is  given. 


TABLE  59 

COST  OF  CANALS  IN  ORDINARY  EARTH 
1910-12 


CAPACITY, 
SEC.  FT. 

AREA  WET, 
SEC.  SQ.  FT. 

DEPTH  OF 
WATER 

SLOPE  IN  FT. 
PER  MILE 

APPROX.  COST  PER  RUNNING  FT 

Low 

High 

50 

25 

2-5 

4 

$0.375 

$0.75 

IOO 

50 

3-5 

2 

0-75 

1-50 

200 
300 
400 

IOO 

150 
200 

5 
6 

7 

1-5 

1  .0 

0-75 

1-50 
2.25 
3-00 

3.00 

4-50 
6.00 

500 

250 

7 

0-75 

3-75 

7-50 

1000 

500 

IO 

0.50 

7-50 

15.00 

1500 

75° 

12 

0.50 

11.25 

22.50 

2OOO 

IOOO 

12 

o-33 

15.00 

30.00 

3000 

1500 

15 

0.25 

22.50 

45  oo 

In  rock  the  allowable  velocity  is  8  feet  per  second  if  the  canal 
is  lined.  Under  such  conditions  the  following  data  is  given. 
The  costs  may  vary  greatly  with  local  conditions. 


H4 


BASIC  COSTS 


TABLE  60 

COST  OF  CANALS  IN  ROCK 
1910-12 


CAPACITY, 
SEC.  FT. 

AREA  WET, 
SEC.  SQ.  FT. 

WATER, 
DEPTH 

SLOPE  IN  FT. 
PER  MILE 

COST  PER  RUNNING  FT. 

Low 

High 

50 

6.25 

2-5 

40 

$0.32 

$1.28 

100 

12.5 

3-5 

25 

0.63 

2.50 

200 

25.0 

S-o 

16 

1.25 

5.00 

300 

37-5 

6.0 

12 

1.87 

7-50 

400 

50.0 

7 

10 

2.50 

IO.OO 

500 

62.5 

7 

9 

3-25 

13.00 

IOOO 

125.0 

10 

6 

6.00 

24.00 

1500 

175.00 

12 

4-5 

8.75 

35-oo 

2OOO 

250.0 

2 

3-5 

12.50 

50.00 

3000 

375-o 

IS 

3-o 

i8.75 

75-oo 

The    Reclamation    Service    gives    the    following    costs    of 
excavation. 

TABLE  61 

COST  or  EXCAVATIONS 

1910-12 


Low 

High 

Average 

Plowable  with  4  horses 

$o  098 

$1    OO 

$0   18 

Plowable  with  6  horses  
Indurated  material  
Loose  rock 

o.  1225 
o.  29 

O    3? 

2.OO 
2.OO 
3OO 

0.30 
O.6o 

O    7"s 

Solid  rock 

o  60 

r    OO 

2   OO 

Excavation  below  plane  of  saturation   . 

o  20 

1    OO 

I    80 

Solid  rock  below  water  

A     CO 

COST  PER  Cu.  YD. 


The  same  authorities,  the  General  Electric  Company,  give 
the  following  valuable  data  on  hydro-electric  installations. 


HYDRO-ELECTRIC  INSTALLATIONS 


.  I 

0  £ 

w  ** 

I-H  PM 

cq  o 

<*  ^ 


^        O         »0 


•   Q 

«o  O 
I       • 

ON  tO 


ON        t^        rf         |  |       • 

M  M  04  t>»          -<*-OO 

.  tf> 


C4O  IOI>. 


•^-  O 

1         vo   0 

10.      O        10       »o       o        10      T     • 

HMMCS  -CSt>. 


O   O 


I  i 

O\MD 


10    68       88 


VO  »O    M 


'o        O 

MM 


0  O    O 

t>.  ON  O  O    O 

1  1       •  •     • 


o          o  o 

O  0    O 


l  ;S    S 


" 


..T  -Hu<. 


O  O 

NO        cl 


Q   Q  O         O 

O    O  t^       CO 

'  '            • 

<NIO  t^^ 


o       o 


1          IN  «o  *>•  oo 

00        V>  H  t-t  **> 

" 


O    O  O         O 

oo          oo 


I  : 


i 


U     H      H      CJ 


UU 


n6  BASIC   COSTS 

In  the  above 

Low  head,  50  to  200  ft. 
Medium  head,  200  to  600  ft. 
High  head,  600  ft.  and  above. 
Small  capacity,  200  to  1000  kw. 
Medium  capacity,  1000  to  5000  kw. 
Large  capacity.  5000  kw.  and  over. 

The  per  cent  of  total  costs  of  various  items  of  construction 
in  hydro-electric  plants  is  given  as  follows: 

SMALL  Low  HEAD  PLANT 

Hydraulic  work  not  including  power  house.  55  % 

Power  house  building 6  %  1 

Water  wheels '. . . .   9  %  \  Power  house  fully  equipped  45  % 

Electric  equipment .3°%  J 

100% 

MEDIUM  Low  HEAD  PLANT 

Dam 43%) 

Low  pressure  pipe 20  %  >  Hydraulic  work  less  power 

High  pressure  pipe 3  %  j      house  66% 

Power  house  and  machinery 34  % 

SMALL  MEDIUM  HEAD  PLANT 

Hydraulic  work  not  including  power  house 

building 76     % 

Power  house  building 8     %  j 

Turbines 3  •  5  %  f  Power  house  fully  equipped  24% 

Electric  equipment 12.5     J 

LARGE  MEDIUM  HEAD  PLANT 

Hydraulic  work 38  % 

Power  house  building 10  %  ] 

Hydraulic  machinery 31  %  \  Power  house  fully  equipped  62  % 

Electric  equipment 21  %  j 

LARGE  HIGH  HEAD  PLANT 
Dam 22  %  1  Hydraulic  work  not  including 


Low  pressure  pipe 23%  ,  r  house  fi 

High  pressure  pipe 16  %  J 

Power  house  fully  equipped 39  % 

85.  More  complete  data  on  the  cost  of  buildings,  excava- 
tions, fills,  foundations,  bridges,  tunnels  and  the  like  are  given 
in  Gillette's  "Handbook  of  Cost  Data"  (Myron  C.  Clark 


REFERENCES  117 

Publishing  Co.).  For  .similar  data  on  power  plants,  the 
reader  is  referred  to  Fernold  and  Orrok's  "Engineering  of 
Power  Plants"  (McGraw  Hill  Book  Co.),  Harding  and  Wil- 
lard's  " Power  Plants  and  Refrigeration"  (John  Wiley  &  Sons, 
Inc.),  and  Guldner's  "Internal  Combustion  Motors"  (D.  Van 
Nostrand  Co.). 

The  equations  of  cost  as  given  by  Harding  and  Willard  in 
their  "Power  Plants  and  Refrigeration"  (John  Wiley  &  Sons, 
Inc.,  1917)  are  as  follows: 


118 


BASIC   COSTS 


• 

ill 

e 

6        .      • 

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• 

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124  •  BASIC   COSTS 


PROBLEMS 

1.  A  150  h.p.  engine  has  a  guaranteed  steam  consumption  of  20$  steam 
per  h.p.h.     The  test  shows  that  the  actual  consumption  is  21.6$.     The 
engine  operates  for  3000  hours  per  year  at  full  load  and  costs  $4500  installed. 
The  steam  costs  25  cents  per  1000$  to  produce.    The  life  of  the  engine  is  20 
years.    The  interest  rate  is  6%.    How  much  should  be  deducted  from  the 
purchase  price  to  compensate  the  purchaser  for  the  increased  cost  of  operation 
above  that  guaranteed? 

2.  A  300  h.p.  engine  has  a  guaranteed  steam  consumption  of  20$  per 
h.p.h.    The  test  shows  that  the  actual  consumption  is  only  18.4$.    The 
engine  operates  for  4000  hours  per  year  at  full  load,  and  costs  $8500  installed. 
The  steam  costs  30  cents  per  iooo#  to  produce.    The  life  of  the  engine  is 
25  years.    Interest  rate  is  7%.    How  much  should  the  manufacturer  receive 
to  compensate  for  the  added  economy  of  service  that  he  has  produced? 

3.  A  power  plant  consists  of  two  5000  h.p.  engines  that  consume  20$ 
of   steam  per  h.p.h.     Boiler  feed  temperature  72°,  steam  pressure  160$ 
and  quality  98%.    The  boilers  evaporate  8.5$  of  water  per  pound  of  coal. 
The  engines  operate  for  3600  hours  per  year  at  full  load.    Interest  rate  is 
6%.    If  these  engines  are  removed  and  sold  for  $6  per  h.p.,  and  new  ones 
are  installed,  having  a  life  of  25  years  and  using  only  12 f  of  steam  per 
h.p.h.,  how  much  can  be  paid  for  the  new  engines?    Heat  costs  18  cents  per 
million  B.t.u. 

4.  A  200  h.p.  automatic  engine  uses  31$  steam  per  h.p.h.  and  costs 
$2800.    A  200  h.p.  Corliss  engine  uses  25$  steam  per  h.p.h.  and  costs  $4200. 
Steam  costs  25  cents  per  iooo#  to  produce.    The  plant  runs  for  3000  hours 
per  year  at  full  load.    Interest  rate  is  5  %.    The  life  of  the  high  speed  engine 
is  15  years  and  for  the  low  speed  engine  it  is  25  years.    Which  engine  is  the 
most  economical?    How  much  saving  will  this  represent  per  year? 

6.  In  problem  (4)  if  the  engines  are  to  be  used  for  only  1000  hours  per 
year  at  full  load,  which  engine  will  be  the  best?  How  much  saving  will 
this  represent  per  year? 

6.  In  problem  (4),  for  what  number  of  hours  of  service  per  year  are  the 
two  engines  equal? 

7.  A  steam  plant  is  to  be  installed  to  deliver  1000  kw.  at  the  switchboard. 
A  steam  engine-generator  set  can  be  installed,  with  a  guaranteed  economy 
of  15$  of  steam  per  i. h.p.h.  with  a  steam  pressure  of  150$,  feed  temperature 
1 80°,  quality  97.5%.    The  mechanical  efficiency  of  the  engine  is  guaranteed 
at  92%,  and  the  generator  efficiency  at  94%.    Or  a  steam  turbo-generator 
can  be  installed  with  a  guaranteed  steam  consumption  of  2o#  per  kw.hr. 
with  a  steam  pressure  of  150$,  feed  temperature  150°  and  150°  of  superheat. 
The  cost  of  generating  steam  in  either  case  is  20  cents  per  1,000,000  B.t.u. 


PROBLEMS  125 

The  steam  engine-generating  set  costs  $23,500.  The  turbo-generator  set 
costs  $19,700.  Superheaters  cost  $1800,  and  boilers  cost  $8.52  per  boiler 
h.p.  The  attendance  cost  is  o.i  cent  per  h.p.h.  for  the  engine-generator 
and  0.06  cent  per  h.p.h.  for  the  turbo-generator.  Interest  rate  is  5%. 
If  the  plant  is  to  be  run  at  full  load  for  6000  hours  per  year,  which  outfit 
is  the  best,  and  by  how  much?  What  is  the  cost  per  kw.h.  delivered? 

8.  If  in  problem  (7),  the  plant  is  to  be  used  for  1000  hours  per  year, 
then  which  is  the  best  and  by  how  much?    What  is  the  cost  per  kw.h. 
delivered? 

9.  In  problem  (7)  for  what  number  of  hours  of  service  per  year  are  the 
two  units  of  equal  worth? 

10.  A  boiler  plant  develops   500  bo.h.p.  and  uses  3$  coal  per  bo.h.p. 
The  steam  pressure  is  150$  and  feed  temperature  80°.    A  feed  water  heater 
is  added  which  costs  $600  and  raises  the  feed  temperature  to  180°.     The 
plant  operates  for  4000  hours  per  year.     Coal  contains  12,725  B.t.u.  per 
pound  and  costs  $4  per  ton.    Interest  rate  is  5%.    How  much  saving  per 
bo. h.p.h.  will  be  effected? 

11.  If  in  problem  (10),  instead  of  a  feed- water  heater,  an  economizer  is 
added,  raising  the  feed  temperature  to  300°,  allowing  7  %  for  depreciation, 
repair  and  attendance,  how  much  could  we  afford  to  pay  for  the  economizer? 

12.  In  problem  (10),  what  would  be  the  saving  per  bo.h.p.  if  the  plant 
were  run  at  full  load  for  only  2000  hours  per  year? 

13.  In  problem  (n),  what  could  we  afford  to  pay  for  the  economizer,  if 
the  plant  were  run  at  full  load  for  only  1000  hours  per  year? 

14.  In  problem  (n),  how  much  could  we  afford  to  pay  for  the  economizer 
if  it  were  used  at  full  load  for  only  TOO  hours  per  year?    For  no  hours 
per  year? 

15.  A  boiler  plant  operates  at  full  load  for  7000  hours  per  year.    It  burns 
50  tons  of  coal  per  day  costing  $4  per  ton  with  operator  (A)  in  charge. 
An  analysis  of  the  flue  gases  shows  CO2,  5  per  cent;  O,  15  per  cent;  and  N, 
75  per  cent.    The  coal  contains  C,  80  per  cent,  H,  6  per  cent,  and  O,  4 
per  cent.    Stack  temperature  600°;  boiler  room  temperature  70°. 

Operator  (B)  is  placed  in  charge  and  another  analysis  taken.  This  shows 
with  the  same  fuel,  C02,  14  per  cent;  O,  6  per  cent;  and  N,  75  per  cent. 
Stack  temperature  520°. 

If  operator  (A)  receives  $1200  per  year,  what  can  we  afford  to  pay  operator 
(B)? 

16.  If  in  problem  (15),  the  analysis  showed  CO2,  13  per  cent,  CO  i  per 
cent,  and  all  other  things  remained  the  same,  what  then  could  we  afford 
to  pay  operator  (B)? 

17.  If  in  problem   (15),  the  analysis  showed  CO2,  10%,  COi,  0.2  per 
cent  and  O,   10%,  with   a    stack    temperature   of    550°,  while   all   other 
things  remain  as  before,  what  then  could  we  afford  to  pay  operator  (B)? 


,  CHAPTER  V 
>      VESTANCES 

Total  Vestance.  The  Time  Element.  Change  Point.  Vestances  of 
Steam  Engines  of  Various  Types  at  Full  and  Fractional  Loads. 
Vestances  at  Full  and  Fractional  Loads  of  Oil  Engines.  Diesel 
Engines.  Induction  Motors  and  Generators.  Comparison  of  Power 
Units  of  Various  Size  and  Types  at  Full  and  Fractional  Loads.  Ves- 
tances at  Full  and  Fractional  Loads  of  Centrifugal  Pumps  and  of 
Standard  and  Wood  Pipe  at  Varying  Capacities. 

86.  As  previously  shown,  the  total  vestance  gives  us  a  basis 
of  comparing  two  or  more  units,  or  two  or  more  systems  that 
are  designed  to  render  similar  service  but  each  differing  from 
the  other  in  one  or  more  of  their  cost  elements,  such  as  first 
cost,  life,  efficiency,  attendance  and  the  like.  The  question 
that  both  the  manufacturer  (producer)  and  user  must  face  con- 
tinually is  one  of  economy.  The  user  desires  to  purchase  that 
unit  which  will  give  him  the  desired  service  at  the  least  ulti- 
mate cost.  The  manufacturer's  problem  leads  to  the  same 
end,  namely  to  produce  a  unit  which  will  represent  the  best 
possible  investment  for  the  purchaser.  But  the  purchaser's 
needs  are  variable.  What  may  be  best  for  one,  may  be  a  very 
poor  investment  for  another  purchaser,  whose  conditions  may 
be  entirely  different.  It  is  therefore  necessary  to  produce 
various  classes  and  types  of  apparatus,  to  meet  these  variable 
conditions  so  that  a  best  investment  for  each  condition  may  be 
offered.  That  is  the  problem  in  general. 

Specifically  the  producer  may  face  such  a  problem  as  this: 
a  more  efficient  engine  can  be  built,  but  at  a  greater  first  cost. 
If  produced,  will  this  engine  represent  a  better  investment 
than  the  unit  now  being  produced  and  will  it  sell?  Whether 
thex  new  engine  will  represent  a  better  investment  for  any 
given  service,  can  be  readily  determined  by  comparing  the  total 
Vestances.  If  that  of  the  new  engine  is  the  smaller,  greater 

126 


THE   TIME   ELEMENT  127 

financial  efficiency  has  been  attained.  Whether  the  engine 
will  sell,  depends  on  whether  it  is  really  a  better  investment  or 
not,  and  by  how  much,  and  second  whether  we  can  prove  to 
the  user  that  it  is  a  better  investment.  Of  course  if  we  can 
prove  it  for  ourselves,  we  can  prove  it  to  the  user,  but  when 
it  comes  to  a  matter  of  guess,  one  man's  guess  is  quite  as  good 
as  another's.  At  all  times  we  must  bear  in  mind  that  we  do 
not  purchase  the  machine,  but  the  service  that  it  will  render 
us.  Manufacturers  therefore  produce  service  in  just  the  same 
sense  that  the  so-called  public  utilities  do. 

87.  The  Time  Element  in  Vestance.—  The  total  vestance 
depends  upon  the  amount  of  use  that  is  made  of  the  apparatus. 
This  may  be  shown  as  follows: 

The  total  vestance  is 


neglecting  insurance  and  taxes,  where 
C  =  the  first  cost, 
T  =*  the  term  factor, 
R  =  the  interest  rate, 
and  A  =  the  annual  cost  of  operation. 

The  annual  cost  of  operation  (^4)  varies  with  the  number  of 
hours  of  use  per  year  and  the  cost  of  operation  per  hour.     If 
a  =  the   mean   operating   cost   per   hour    of    the    entire 

unit  or  system  and 

N  =  the  number  of  hours  of  use  per  year,  then 
A  =  aN 
and  the  total  vestance  becomes 

'     .         '     J         V    -£  +  !£.       .......     (64) 

Dividing  equation  (16)  through  by  (N)  gives  the  vestance 
per  hour  Vn  of  service  or 


By  equation  (65)  it  is  evident  that  as  (N),  the  number  of 
hours  of  use  increases,  the  vestance  per  hour  decreases.     The 


128  VESTANCES 

maximum  value  that  (N)  can  attain  is  8760,  the  total  num- 
ber of  hours  in  a  year.  For  this  value  of  N,  the  vestance  per 
hour  of  service  is  a  minimum.  When,  on  the  other  hand,  ( N) 
decreases,  the  vestance  per  hour  increases,  becoming  infinite 
when  N  =  o. 

Evidently  then  as  ( N)  becomes  smaller,  and  the  term  (  — —  J 
becomes  larger,  the  constant  term  (  —  )  becomes  of  less  and 


less  consequence.  We  can  interpret  this  for  practical  use  by 
saying  that  where  the  unit  is  but  little  used  per  year,  the  first 
cost  is  of  primary  consideration,  and  the  operating  cost  per 
hour  (efficiency  and  energy  cost)  of  secondary  importance 
only,  while  for  many  hours  of  operation  per  year  conditions 
are  reversed.  That  is,  when  the  apparatus  is  much  used,  we 
can  afford  to  buy  expensive  equipment  in  order  to  get  low 
cost  of  operation,  but  where  little  used,  cheap  apparatus  of 
'comparatively  low  efficiency  is  the  best. 

88.   Valuance.  —  If  we  take  the  reciprocal  of  the  vestance 
per  hour  (Vh),  we  get  by  equation  (65) 


TT  X 

"=" 


TRN 


h       CR+TaN      •••• 

We  shall  call  (U)  the  valuance.  It  is  a  measure  of  the 
worth  of  the  unit.  When  (N),  the  number  of  hours  of  use,  is 
zero,  the  valuance  is  zero,  i.e.,  the  unit  is  worthless  (useless). 
The  valuance  increases  as  (N)  increases,  becoming  a  maxi- 
mum when  N  =  8760,  i.e.,  the  apparatus  is  worth  most  (most 
useful)  when  it  is  used  continually. 

89.  We  shall  now  determine  the  vestances  of  various  types 
of  equipment  for  a  normal  year  of  3000  hours,  assuming  a 
constant  load  during  the  entire  period.     In  this  way  we  can 
get  a  first  approximation  of  the  comparative  values  of  such 
equipment.     Further  on,  we  shall  consider  more  variable  and 
exact  conditions. 

90.  The  Steam  Engine.  —  From  the  previous  chapter,  we 
find  that  the  first  cost  of  a  100  h.p.  low  speed  compound  steam 


THE  STEAM  ENGINE  129 

engine  is  $3300.  The  efficiency  with  dry  saturated  steam  at 
ioo#  initial  steam  pressure  is  12  %.  The  cost  of  oil,  attendance, 
etc.,  is  about  $0.05  per  hour  for  this  size  of  engine,  and  the 
maintenance  about  $0.02  per  hour.  The  life  of  low  speed 
compound  engines  is  25  years. 

If  the  interest  rate  is  5%,  then  by  Table  2,  the  term  factor 
is  0.70469,  so  that  the  depreciation  vestance  is 

Vd  =    33°°    =  $4682.90. 
0.70469 

If  the  coal  used  for  fuel  contains  13,743  B.t.u.  per  pound, 
its  energy  content  in  h.p.h.  is 


2545 

If  this  coal  costs  $2  per  ton,  or  $0.001  per  pound,  then  its 
cost  per  h.p.h.  is 

o.ooi  -s-  5.4  =  $0.000185. 

If  the  combined  efficiency  of  the  boiler,  furnace,  and  grate 
is  70%,  then  the  cost  per  h.p.h.  of  heat  energy  delivered  to  the 

engine  is 

0.000185  -r-  0.7  =  $0.000264. 

To  this  we  must  add  the  other  costs  of  producing  steam, 
amounting  to,  let  us  say,  $0.000236,  giving  a  total  cost  of  the 
heat  delivered  to  the  engine  of 

0.000264  +  0.000236  =  $0.005  per  h.p.h. 
Since  the  efficiency  of  the  engine  is  12%,  the  cost  of  the  heat 
per  i.h.p.  is 

0.0005  -4-  o.i  2  =  $0.00417, 

and  for  100  h.p.  this  amounts  to 

0.00417  X  100  =  $0.417. 

To  this  must  be  added  the  other  operating  costs  of  $0.07, 
giving  a  total  of 

0.417  +  0.07  =  $0.487  per  hr. 

For  a  year  of  3000  hours,  this  amounts  to 
0.487  X  3000  =  $1461, 

giving  an  operating  vestance  at  5  %  interest,  of 
1461  -T-  0.05  =  $29,220. 


130  VESTANCES 

Since  the  depreciation  vestance  is  $4682,  the  total  vestance 
is 

Vt  =  4682  +  29,220  =  $33,902, 
or  $339.02  per  i.h.p. 

Note  that  the  operating  vestance  is  over  six  times  as  great 
as  the  depreciation  vestance.  In  other  words  the  first  cost  is 
of  relatively  small  importance  as  compared  with  the  costs  of 
operation,  a  matter  not  always  fully  appreciated. 

91.   As  compared  with  the  above,  let  us  determine  the  total 
vestance  of  the  above  engine  when  the  initial  pressure  of  the 
steam  is  2oo#,  other  things  remaining  the  same. 
Since  for  compound  engines,  the  efficiency  is 

y  =  0.0675  (£  ~  i45)i 
and  the  steam  temperature  (£)  at  2oo#  is  388°,  we  find  that 

y  =  0.0675  (388  -  145)  =  16.4% 
instead  of  12%  previously  taken. 
The  heat  will  cost  therefore 

0.0005  -f-  0.164  =  $0.00305  per  h.p.h., 
or  0.00305  X  100  =  $0.305  per  hr., 

giving  a  total  cost  per  hour  of 

0.305  +  0.07  =  $0.375, 

or  O-375  X  3000  =  $1125  per  normal  year. 

The  operating  vestance  is  then 

1125  -T-  0.05  =  $22,500, 
giving  a  total  vestance  of 

22,500  +  4682  =  $27,182. 
This  shows  a  reduction  of 

33,902  -  27,182  =  $6720, 
or  $67.20  per  h.p. 

If  we  add  to  the  2oo#  initial  gage  pressure  200°  of  super- 
heat, then  the  efficiency  will  be  increased  by  20%,  amounting 
to 

1.20   X    16.4    =    19. 

The  cost  of  heat  per  hour  will  then  be 


THE  STEAM  ENGINE  131 

and  the  total  cost  per  hour  will  be 

0.254  +  0.07  =  $0.324, 
or  0.324  X  3000  =  $972, 

giving  an  operating  vestance  of 

972  -f-  0.05  =  $19,440, 
and  a  total  vestance  of 

•4682  +  19,440  =  $24,122. 
This  represents  a  saving  of 

27,182  —  24,122  =  $3060, 

as  compared  with  the  same  unit,  not  using  superheat  but  trie 
same  pressure,  and 

33,902  -  24,122  =  $9780, 

as  compared  with  the  engine  using  dry  saturated  steam  at 
ioo#  initial  pressure. 

The  per  cent  saving  by  use  of  2oo#  pressure  as  compared 
with  ioo#  initial  gage  pressure  is 

6720  ^  33,902  =  20%. 

The  saving  in  total  vestance  by  the  use  of  2oo#  pressure 
together  with  200°  superheat  as  compared  with  the  use  of  ioo# 
dry  steam  pressure  is 

978°  +  33.902  =  28.8%. 

Note  the  enormous  saving  especially  with  the  use  of  higher 
pressures.  Modern  practice  tends  towards  continually  higher 
pressures.  That  this  is  good  practice  is  too  clearly  shown 
above  to  need  further  emphasis. 

92.  Let  us  compare  with  the  above  the  total  vestance  of  a 
1000  h.p.  compound  engine,  the  life,  rate  of  interest  and  cost 
of  coal  being  assumed  the  same  as  that  for  the  100  h.p. 
engine. 

For  this  engine  the  first  cost  is  $18.60  per  h.p.  and  the  de- 
preciation vestance  is  then 

18.60  -f-  0.70469  =  $26.394  per  i.h.p. 

At  ioo#  initial  gauge  pressure,  and  dry  saturated  steam,  the 
efficiency  of  the  engine  is  14.7%.  The  heat  cost  per  h.p.h. 
will  then  be 

0.0005  •*•  °-I47  =  $0.0034. 


I32  VESTANCES 

Allowing  $0.0006  as  the  cost  of  attendance,  maintenance, 
etc.,  per  h.p.h.,  the  total  operating  cost  per  h.p.h.  will  be 

0.0034  +  0.0006  =  $0.004, 

or  0.004  X  3000   =  $12  per  h.p.  year. 

The  operating  vestance  is  then 

12-f-  0.05  =  $240  per  h.p., 
and  the  total  vestance  per  h.p.  year  is 

240  +  26.394  =  $266.394, 

as  compared  with  a  total  vestance  of  339.02  per  h.p.h.  for  the 
100  h.p.  engine.     This  gives  a  difference  of 

339.02  -  266.394  =  $72.626. 

The  per  cent  saving  as  compared  with  the  100  h.p.  engine  of 
72.626  -T-  339.02  =  21.4%. 

This  is  of  course  not  a  clear  saving  due  to  the  greater  dis- 
tribution area,  and  corresponding  greater  distribution  cost  for 
the  larger  plant.  In  fact  the  vestance  of  this  increased  cost 
of  distribution  may,  if  the  system  is  too  large,  more  than  off- 
set the  decreased  vestance  of  the  engine  itself. 

93.  We  shall  now  determine  the  total  vestance  of  a  100  h.p. 
low  speed  simple  noncondensing  engine  operating  on  dry 
saturated  steam  at  ioo#  initial  gage  pressure,  the  life,  cost 
of  coal,  and  interest  rate  remaining  the  same  as  above. 

The  first  cost  of  such  an  engine  is  $21.50  per  h.p.  The 
depreciation  vestance  is  therefore 

21.50  -r-  0.70469  =  $30.51  per  h.p. 

This  engine  will  use  42%  more  steam  than  a  compound 
condensing   engine.      The   efficiency   of   the   simple   noncon- 
densing engine  of  100  i.h.p.  will  therefore  be 
12  -T-  1.42  =  8.45%. 

The  cost  of  the  heat  energy  per  h.p.h.  will  therefore  be 
0.0005  -7-  0.0845  =  $0.0059. 

Allowing  as  before  $0.0007  for  the  other  operating  costs,  we 
get  a  total  of 

0.0059  +  0.0007  =  $0.0066  per  h.p.h., 
or  0.0066  X  3000  =  $19.80  per  h.p.  year. 


THE  STEAM  ENGINE  133 

The  operating  vestance  is  then 

19.80  -f-  0.05  =  $396, 

and  the  total  vestance  is 

396  +  30.51  =  $426.51, 

as  compared  with  339.02  for  the  100  h.p.  condensing  engine. 
The  difference  is 

426.51  -  339.02  =  $87.49, 
or  87.49  •*•  339.02  =  26.8%. 

94.  For  a  100  i.h.p.  simple   noncondensing  high  speed  en- 
gine operating   on  dry  saturated  steam  at  ioo#   initial  gage 
pressure,  the  first  cost  is  $12.75  per  h.p.  but  the  life  now  is 
only  15  years  instead  of  25  years  for  the  low  speed  engines. 
The  depreciation  vestance  is  then 

12.75  •*•  0.51897  =  $24.58. 

Since  the  simple  high  speed  noncondensing  engine  will  use 
64%  more  steam  than  a  compound  condensing  engine  and  the 
efficiency  of  a  100  i.h.p.  engine  of  the  latter  type  is  12%,  the 
efficiency  of  the  high  speed  engine  will  be 

12  -T-  1.64  =  7.3%.  . 

The  cost  of  the  heat  energy  per  h.p:h.  is  then 
0.0005  -T-  0.073  =  $0.00685. 

Adding  thereto  the  other  operating  cost  per  h.p.h.  we  get  for 
the  total  operating  cost  per  h.p.h. 

0.00685  +  0.0007  =  $O-OO755, 

or  0.00755  X  3000  =  $22.65  Per  h-P-  year. 

The  operating  vestance  is  therefore 

22.65  +  °-°5  =  $453, 
and  the  total  vestance  is 

453.00  +  24.58  =  $477-58, 

as  compared  again  with  339.02   for  the  compound  engine. 
This  gives  a  difference  of 

477-58  -  339-02  =  $138.56, 
or  138.56  -5-  339-02  =  41%. 

95.  In  order  to  bring  out  the  differences  more  clearly,  we  tabu- 
late the  vestances  of  the  three  classes  of  100  i.h.p.  engines  below. 


134 


VESTANCES 


TABLE  64 

VESTANCES 

100  I.H.P.  ENGINES  WITH  DRY  SATURATED  STEAM  AT  IOO#  G.P. 


Engine 

Vd 

Va 

F 

Percent  Vt. 

Va 

Vd 

A 

46.82 

292.  2O 

339.02 

I  .OOO 

6.23 

B 

30-51 

396  .  oo 

426.51 

1.268 

13.00 

C 

24.58 

453-oo 

477-58 

1  .410 

18.40 

Here  engine  A  =  low  speed  compound  condensing;  B  =  low 
speed  simple  noncondensing;  and  C  =  high  speed  simple  non- 
condensing. 

Note  now  that  between  engines  A  and  B  we  save  $16.31 
in  depreciation  vestance  at  a  cost  of  $103.80  in  operating  ves- 
tances, and  between  A  and  C  we  save  $22.24  in  deprecia- 
tion vestance.  Neither  of  these  represent  very  good  bargains. 
Yet  how  often  are  such  bargains  entered  into! 

96.  We  tabulate  below  the  vestances  of  steam  engines, 
based  on  5%  interest,  $2.50  cost  per  ton  of  coal  of  fuel  value 
as  above,  i.e.,  5.4  h.p.h.  per  pound  and  25  years'  life  for  low 
speed  engines,  and  15  years  for  high  speed  engines,  all  as  taken 
above.  The  effect  of  increased  cost  of  operation  is  to  increase 
the  operating  vestance  by  a  similar  amount.  Thus  if  coal  were 
$5  per  ton  and  other  operating  costs  increased  in  like  pro- 
portion, the  operating  vestance  would  be  doubled,  amounting 
to  some  $400  increase.  If  on  the  other  hand  the  first  cost  is 
doubled,  the  depreciation  vestance  will  be  doubled,  but  this 
will  amount  only  to  some  $30.  Evidently  then  the  effect 
of  increased  prices  is  to  make  efficiency  relatively  more  im- 
portant and  first  cost,  though  itself  increased,  of  relatively  less 
importance. 

The  effect  of  increased  interest  rate  is  to  decrease  the 
vestances. 


THE  STEAM  ENGINE 


135 


TABLE  65 

VESTANCES  PER  H.P.  or  COMPOUND  CONDENSING  Low  SPEED 
STEAM  ENGINES  AT  FULL  LOAD 


VESTANCES 

RATIO 

SIZE, 

I.H.P. 

Depreciation  Vd 

Operating  Va 

Total  V 

Va 

Vd 

IOO 

$46.829 

$292.00 

$338-83 

7-2 

*  .            20O 

35.476 

271  .00 

306.48 

8.7 

300 

31.688 

262.00 

293.69 

9-3 

4OO 

29  .  800 

256.00 

285.80 

9.6 

500 

28.736 

251  .  2O 

279.94 

9-7 

600 

27.913 

247  •  40 

275-5I 

9.8 

700 

27.572 

244.  10 

271.67 

99 

800 

26.962 

242.00 

269  .  oo 

10.  0 

QOO 

26.600 

241  .10 

267  .  70 

IO.O 

IOOO 

26.394 

240  .  oo 

266  .  39 

IO.I 

1500 

25.642 

236.00 

261  .  64 

IO.2 

2OOO 

2S-259 

234  .  oo 

259.26 

10.2 

^ 


cv300 


>100 


ting  V 


_Uei 


jrec  ation  Ves 


200        400        600        800       1000       1200       1400       1600       1800     2000 
H.P. 

FIG.  21.  —  Vestances  per  h.p.  of  compound  low  speed  condensing 
steam  engines. 


136 


VESTANCES 


TABLE  66 

VESTANCES  PER  H.p.  OF  SIMPIE  Low  SPEED  NON CONDENSING 
STEAM  ENGINES 


VESTANCE  AT  FULL  LOAD 

RATIO 

SIZE, 

I.H.R 

Depreciation  Vd 

Operating  Va 

Total  V 

Va 

vd 

IOO 

$30-SIO 

$398  .  oo 

$428.51 

J3-i 

125 

28.381 

386  .  oo 

414.38 

13-7 

ISO 

26.962 

382.00 

408  .  96 

14.2 

175 

25.898 

374-oo 

399.90 

14.4 

200 

25.188 

368  .  oo 

393-19 

14.6 

250 

24.124 

362  .  oo 

386.12 

15.0 

300 

23-4I5 

358.00 

381-41 

J5-3 

350 

22.918 

354-oo 

376.92 

J5-4 

400 

22.521 

350.00 

372.52 

15-5 

450 

22.237 

346  •  oo 

368.24 

iS-6 

500 

21.996 

342.00 

364.00 

15-5 

TABLE  67 

VESTANCES  PER  H.P.  OF  SIMPLE  HIGH-SPEED  NONCONDENSING 
STEAM  ENGINES 


VESTANCE  AT  FULL  LOAD 

RATIO 

SIZE, 

I.H.P. 

Depreciation  Vd 

Operating  Va 

Total  V 

Va 

Vd 

20 

$49-3l3 

$874  .  oo 

$923-3i 

17.7 

30 

36.328 

618.00 

654-33 

17.0 

40 

29  .  800 

550.00 

579.80 

18.5 

50 

25-898 

503  oo 

528.90 

19.4 

75 

20.718 

474.00 

494.72 

22.9 

IOO 

18.093 

454-oo 

472.09 

25-0 

150 

15.462 

434-00 

449.46 

28.5 

200 

14.191 

418.00 

432.19 

29.6 

250 

13-339 

41  2  .  OO 

425-34 

31-4 

THE  STEAM  ENGINE 


137 


, 

I 

l\ 

\ 

^ 

-- 

= 

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—  —  - 

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•    - 

—  — 

•    ^ 

•     • 

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mplt 

=^—  - 

Hig 

2om] 

^9^1M 

d-  sp< 
»ounc 

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on-co 
jle-L 
-spee 

ndea 

3W--SI 

•^.HM 

sing 
eed._ 

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—  — 

Ingi 
Sons 

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_^__ 

ies 
onde 

—. 

nsint 

~ 

^_« 

JEng 

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ines. 

S^ 

0          50         100        150        200        250        300        350        400        450       50( 
H.  P. 

$1000 


800 


400 


200 


FIG.  22.  —  Total  vestances  per  kp.  of  steam  engines. 

A  comparison  of  these  tables  shows  that  the  total  vestance 
decreases  with  increased  size  of  unit.  The  decrease  is  rapid 
for  units  less  than  100  h.p.,  rather  small  between  100  and 
500  h.p.,  and  almost  insignificant  beyond  that.  Again  the 
approximate  mean  vestance  ratio  is  10  for  compound  con- 
densing engines,  15  for  simple  low  speed  noncondensing  en- 
gines and  about  25  for  simple  high  speed  noncondensing 
engines.  That  is  as  the  first  cost  is  reduced,  the  relative 
amount  of  money  that  must  be  spent  for  operation  is  enor- 
mously increased.  In  as  much  as  the  vestance  is  proportional 
to  the  total  amount  of  labor  that  it  takes  to  produce  a  given 
service,  including  not  only  the  labor  of  operation  but  of  the 
construction  of  the  units  and  the  system,  it  appears  that  in 
cheap  layouts,  too  little  effort  has  been  spent  in  the  design 
and  construction  of  the  units  and  system  and  too  much  must 
therefore  be  spent  in  operation.  Less  total  labor  would  be 
required  if  it  were  used  in  the  right  place,  in  properly  prepar- 
ing for  the  rendition  of  the  service  first,  before  rendering  it. 
On  the  other  hand,  too  much  may  be  spent  on  preparation. 

97.  The  above  tables  were  figured  on  the  basis  of  a  normal 
year  of  3000  hours  of  use.  If  the  number  of  hours  of  use  are 
changed,  then  the  cost  of  operation  is  changed,  since  the  latter 


138  VESTANCES 

is  in  direct  proportion  to  the  hours  of  use.     But  the  deprecia- 
tion vestance  remains  constant.     We  have  then,  that 

F=  Fz>  +  Va'N, (67) 

where 

Va   =  the  operating  vestance  per  hour, 
and       N  =  the  hours  of  use, 

VD  =  the  depreciation  vestance, 
and       V  =  the  total  vestance. 

From  the  above,  it  is  evident  that  the  vestance  per  hour  is 

Vk=¥f+Va> (68) 

Example  27.  —  Determine  the  total  vestance  per  hour  of  a 
500  h.p.  compound  condensing  steam  engine  for  a  variable 
number  of  hours  of  use  per  year. 

Solution:  By  Table  65,  we  have 

'  V    =  $28.736, 
and  Va  =  $251.20  for  N  =  3000  hrs. 

Then  F0'  =  **£*°  =  $0.08373. 

3000 

Therefore  the  total  vestance  per  hour  is 
Vj  =  ^  +  0.08373. 

From  this,  we  tabulate  as  follows: 

N  Vh 

5000  hrs.  $0.08948 

1000  hrs.  0.11247 

100  hrs.  0.3711 

i  hr.  28.82 

ohr.  oo 

The  total  vestance  of  this  engine  is 

F  =  28.736  +  0.08373  N. 

Example  28.  —  Determine  likewise  the  total  vestance  per 
h.p.  for  (a)  a  100  h.p.  compound  condensing  engine,  (b)  for  a 
100  h.p.  simple  low  speed  noncondensing  engine  and  (c)  for 
a  100  h.p.  simple  high  speed  noncondensing  engine. 


THE   STEAM   ENGINE 

Solution:   (a)  In  this  case,  we  have  by  table  64 
„  _  46.829       292.00 

Vh        ~~  '' 


so  that 


Vh"  = 


+  0.09733, 


V"  =  46.829  +  0.09733 
In  this  case  we  have  by  Table  65 

Vh'"  =  ^ 


-22*12  +  0.13*67, 


i\  3000 

^OXIO    . 

or 

so  that  V"  =  30.510  +  0.13267  N (b) 

(c)  In  this  case,  we  have  by  Table  66 

T/  „„       18.093   , 

I/       '"'         S*-'  I—      f*.      -T    £  T   <}   1 

**•               -ft          °-I5I33> 
or  V""  =  18.093  +  0.15133  N (c) 

Plotting  these  equations  (a),  (6),  and  (c)  gives  straight  lines 
as  shown  in  Fig.  23. 

It  may  be  that  for  a  certain  number  of  hours  of  use  per 

$280 


& 


200 


&w/ 


\T& 


1 160 

I 

3 120 
I 


40 


200 


400         600 


800        1000       1200       1400       1600       1800      2000 
Hours  of  Use  per  Year 


FIG.  23.  —  Variation  of  total  vestance  with  hours  of  use  for  100  h.p. 
steam  engines. 


140  VESTANCES 

year,  that  the  total  vestance  for  the  compound  and  simple 
low  speed  engines  are  equal.     In  such  a  case  then 

V"  =  V", 

or          46.829  +  0.09733  #1  =  3°-510  +  0.13267  Ni, 
whence  Ni  =  461.1  hrs. 

So  again  the  total  vestance  per  year  (VSL}  for  the  simple 
slow  speed  engine  is 

VSL  =  3°-510  +  0.13267^, 
and  for  the  simple  high  speed  engine  it  is 

VSH  =  18.093  +  0.15133^. 
These  two  become  equal  when 

18.093  +  0.15133  N2  =  3°-510  +  0.13267^2, 
that  is  when  Nz  =  661.5  hrs. 

So  again  the  vestance  of  the  compound  engines  equals  that 
of  the  simple  high  speed  engine  when 

18.093  +  0.15133^3  =  46-829  +  0.09733^8, 
or  Ns  =  532.1  hrs. 

98.  Change  Points.  —  The  meaning  of  these  three  points 
is  clearly  shown  in  Fig.  23  where  the  total  vestance  per  horse 
power  year  has  been  plotted  against  the  hours  of  use.  The 
equations  for  vestance  in  terms  of  hours  of  use  give  straight 
lines  as  shown. 

From  this  it  is  evident  that  when  the  hours  of  use  are  less 
than  461.1,  then  the  simple  low  speed  engine,  in  spite  of  its 
lower  efficiency,  is  more  economical  in  use,  i.e.,  has  better 
financial  efficiency  than  the  compound  engine.  When  used  for 
more  than  461.1  hours,  however,  the  compound  engine  is  best. 

So  also  when  the  hours  of  use  are  less  than  532.1,  the  simple 
high  speed  engine  has  better  financial  efficiency  than  the  com- 
pound engine.  But  this  number  of  hours  is  greater  than  that 
of  the  simple  low  speed  engine.  So  then  for  more  than  532.1 
hours  per  year  use  the  compound  engine;  for  less  than  this 
number  of  hours  use  the  simple  high  speed  engine  and  do  not 
use  the  simple  low  speed  engine  at  all.  This  is  for  the  reason 
that  for  no  hours  of  use  during  the  year  is  it  best,  being  always 
exceeded,  either  by  the  compound  or  simple  high  speed  engine. 


CHANGE  POINTS  141 

We  shall  call  the  point  of  equality  as  NI,  Nt,  and  Na  above 
the  Change  Points. 

99.  For  the  200  i.h.p.  engines,  we  have  a  total  vestance  for 
the  compound  engine  of 

V*  =  35476  +  0.09033^, (a) 

for  the  simple  low  speed  engine  of 

VSL  =  25.188+0.12267^,    ......   (b) 

and  for  the  simple  high  speed  engine  of 

VSH  =  14-iQ1  +  0-13733^ (c) 

Equating  equations  (i)  and  (2)  gives 

25.188  +  0.12267  #1  =  3S-476  +  0.09033^1, 
so  that  Ni  =  318  hrs., 

and  so  also  equating  equations  (i)  and  (3)  gives 

14.191  +  0.13733  #2  =  35-476  +  0.09033^2 
so  that  N2  =  452  hrs. 

Again  the  value  of  the  change  point  is  greater  for  the  high 
speed  than  for  the  low  speed  engine  with  the  compound  engine. 
So  again  we  find  that  the  simple  low  speed  engine  possesses 
no  period  of  superiority  and  should  therefore  under  •  the  condi- 
tions assumed  not  be  used  for  this  size.  For  this  size  and  these 
conditions,  the  compound  engine  should  be  used,  if  the  period  of 
use  is  over  452  hours,  and  when  the  period  of  use  is  less  than 
this  number  of  hours,  then  the  simple  high  speed  engine  gives 
the  highest  financial  efficiency. 

100.  So  far  we  have  discussed  only  the  condition  of  an 
engine  operating  at  full  load.  If  instead  the  engine  is  used 
at  a  constant  fractional  load,  the  efficiency  will  be  decreased 
and  thereby  the  cost  of  heat  will  be  increased.  The  other 
operating  costs  such  as  attendance  and  the  like  will  be 
changed  to  correspond  to  the  actual  load  carried.  The  depre- 
ciation vestance  will  change  in  inverse  proportion  to  the  per 
cent  of  full  load  carried.  Thus  at  half  load  it  will  be  doubled; 
at  one  and  one-quarter  load  it  will  be  four-fifths  as  great,  etc. 
This  is  evident  since,  for  example,  at  half  load  we  have  pur- 
chased two  horse  power  to  get  one. 

We  give  below  the  vestances  for  each  of  the  three  types  of 
engines  at  1.25,  0.75,  0.50  and  0.25  of  full  load,  the  per  cents 


142 


VESTANCES 


of  full  load  efficiency  at  fractional  loads  being  assumed  the 
same  for  simple  low  speed  engines  for  want  of  authentic  data 
on  the  former.  The  vestances  are  per  actual  horse  power 
developed  and  not  per  rated  horse  power.  A  normal  year  of 
3000  hours  is  again  assumed. 

TABLE  68 

VESTANCES  PER  H.P  OF  Low  SPEED  COMPOUND  CONDENSING 

STEAM  ENGINES  AT  FRACTIONAL  LOADS 

DRY  SATURATED  STEAM  AT  IOO#  G.P. 


RATED 
H.P. 

TOTAL  VESTANCE  AT  LOADS  OF 

125% 

100% 

75% 

5°% 

25% 

IOO 

$345  •  5 

$338.8 

$361.4 

$41  i  .  6 

$544-3 

200 

313-4 

306.5 

321.3 

361.0 

469.9 

300 

3oi-4 

293-7 

308.3 

345-4 

444-8 

400 

291.8 

285.8 

298.3 

334-2 

424.0 

500 

284.9 

280.0 

291.8 

326.4 

4I5-6 

600 

280.3 

275-5 

286.6 

321.8 

407.6 

700 

278.2 

271.7 

283.0 

318.5 

403-7 

800 

276.6 

269.0 

281.1 

3I5-4 

400.0 

900 

275.0 

267.7 

279.0 

3i3-i 

397-6 

IOOO 

274.0 

266.6 

277.6 

3*1-4 

395-6 

1500 

270.5 

261.6 

273:2 

306.0 

387.2 

2000 

267.5 

259-3 

270.5 

302.2 

383-8 

1800      200 


FIG.  24.  —  Vestances  per  h.p.  of  low  speed  compound  condensing 
steam  engines. 


CHANGE  POINTS 


143 


TABLE  69 

VESTANCES  PER  H.P.  OF  SIMPLE  Low  SPEED  NONCONDENSING 
STEAM  ENGINES  AT  FRACTIONAL  LOADS 

DRY  SATURATED  STEAM  AT  IOO# 


RATED 
H.P. 

TOTAL  VESTANCES  AT  LOADS  OF 

125% 

100% 

75% 

50% 

25% 

IOO 

$442.2 

$429.5 

$442.6 

$496-4 

$609.0 

125 

433-1 

416.2 

431-9 

478.2 

590.6 

150 

424.4 

406.4 

422.O 

466.4 

574-o 

175 

417.1 

398.9 

412.9 

456.8 

562.6 

2OO 

4IO.  2 

393-6 

405.9 

448.4 

554-8 

250 

401.9 

386.1 

395-8 

440.0 

543-0 

300 

396.3 

379-4 

390.1 

434-2 

531-6 

35° 

390-7 

374-7 

384.2 

427.8 

524.6 

400 

384.8 

369-9 

379-6 

421  .6 

5i7.o 

45° 

380.6 

364.6 

375-7 

416.4 

510.4 

500 

375-6 

360.6 

370.9 

412.0 

5°5-4 

TABLE   70 

VESTANCES  PER  H.P.  OF  SIMPLE  HIGH  SPEED  NONCONDENSING 
STEAM  ENGINES  AT  FRACTIONAL  LOADS 

DRY  SATURATED  STEAM  AT  IOO  G.P. 


RATED 


VESTANCES  AT  LOADS  OF 


H.P. 

125% 

100% 

75% 

50% 

25% 

20 

$742.3 

$701.3 

$725.6 

$806.6 

$1005.2 

30 

647.0 

618.3 

638-5 

706.6 

867.2 

40 

587.8 

561.8 

579-8 

639-6 

780.2 

50 

555-7 

529-9 

544-5 

602.8 

731.6 

75 

518.5 

494-7 

506.8 

558.0 

673.6 

IOO 

499.1 

474-1 

485.2 

532.4 

637-4 

150 

474-8 

449.8 

460.2 

504.0 

602.4 

200 

457-4 

432.8 

442.3 

487.4 

574.8 

250 

448.4 

425-9 

434-1 

474-4 

560.2 

144 


VESTANCES 


$1000 


FIG.  25. — Vestances  per  h.p.  of  simple  high  speed  noncondensing 
engines  at  fractional  loads. 

101.  It  will  be  noticed  that  especially  in  the  simple  low 
speed  noncondensing  engines  the  total  vestance  per  h.p.  at 
three-quarters  load  is  but  little  greater  than  at  full  load.  The 
total  vestance  at  half  and  one  and  one-quarter  loads  is  not 
very  great  as  compared  with  that  at  full  load,  while  that  at 
one-fourth  load  is  comparatively  very  great.  The  latter  is 
evidently  then  a  very  uneconomical  load  at  which  to  operate. 

Since  the  above  tables  are  based  on  3000  hours  of  opera- 
tion, we  can  again  get  the  variation  of  total  vestance  with  the 
number  of  hours  of  operation  per  year,  and  thus  determine 
the  change  point  between  two  different  types  of  engines  at  frac- 
tional loads. 

Example  29.  —  Determine  the  change  point  between  a 
100  h.p.  compound  engine  running  at  full  load  and  a  200  h.p. 
compound  engine  running  at  half  load. 

Solution:  The  depreciation  vestance  of  a  100  h.p.  engine  at 
full  load  is  $46.80,  and  the  operating  vestance  is  $292  per  year 
or  292  -s-  3000  =  $0.09733  per  hr. 

The  total  vestance  ( V)  is  then 

V  =  46.80  +  0.09733  N. 


CHANGE  POINTS  145 

For  the  200  h.p.  compound  engine  operating  at  half  load, 
the  depreciation  vestance  is 

35-5  X  2  =  $71.00  per  h.p., 

and  the  operating  vestance  is  the  difference  between  the  total 
and  depreciation  vestances  or 

361  -  71  =  $290  per  yr., 
or  290  -T-  3000  =  $0.09667  per  hr. 

The  total  vestance  is  then 

F20o  =  71  +  0.09667  N. 
The  change  point  is  obtained  by  equating 

Fioo  and  F20o 

or  46.80  +  0.09733  A7"].  =  71  +  0.09667^1, 

or  Ni  =  36,120  hrs. 

Since  there  are  only  8760  hours  in  a  year,  the  change  point 
does  not  come  within  the  year.  At  all  times  then  the  100 
h.p.  engine  is  the  most  economical  to  use  at  100%  load,  the 
decreased  operating  vestance  of  the  200  h.p.  engine  being 
insufficient  to  offset  the  increased  depreciation  vestance  over 
that  of  the  smaller  engine. 

Example  30.  —  Determine  the  change  point  of  a  250  h.p. 
simple  high  speed  engine  running  at  full  load  and  a  200  h.p. 
engine  of  the  same  type  running  at  25%  overload'. 

Solution:  In  this  case  we  have  for  the  250  h.p.  engine 

TV  =  $13.30, 

and  Va   =  412.60  per  yr., 

or  412.6  -r-  3000  =  $0.13753  Per  hr-> 

so  that  VI  =  13.30  +  0.13753  N. 

So  also  for  the  200  h.p.  engine  at  125%  load 

VD"  =  14.2  -f-  1.25  =  $11.40, 

and  Va"  =  4574  -  n-4  =  $446  per  yr., 

or  446  -r-  3000  =  $0.14867  per  hr., 

so  that  V"  =  11.4  +  0.14867  N. 


I46  VESTANCES 

Equating  V I  and  V  "  gives 

11.4  +  0.14867^1  =  13.3  +  0.13753  Ni 
so  that  Ni  =  170  hrs. 

This  shows  that  if  an  overload  of  25%  is  to  be  carried  less 
than  170  hours  per  year,  it  can  be  done  more  economically 
with  the  use  of  the  200  h.p.  engine  at  25%  overload  than  by 
using  a  250  h.p.  engine  at  full  load. 

102.  Oil  Engines.  —  We  give  in  Table  71  the  vestances  per 
h.p.  of  oil  engines  per  normal  year  of  3000  hours,  at  5% 
interest  rate.  These  are  also  based  on  a  fuel  cost  of  0.511 
cent  per  pound  of  oil  having  a  heat  value  of  18,600  B.t.u. 
per  pound.  This  is  equivalent  to  a  heat  value  of  7.3  h.p.h. 
per  pound  and  a  cost  of  heat  energy  of  0.07  cent  per  h.p.h. 
The  life  of  the  engines  is  taken  as  15  years,  although  this 
is  usually  exceeded. 


OIL  ENGINES 


TABLE   71 
VESTANCES  PER  H.p.  OF  OIL  ENGINES 


CT7c. 

LOADS 

H.P. 

TYPE 

100% 

75% 

50% 

25% 

2 

i  cy  horiz 

VD  $174.00 

$231  .  so 

$348  .  oo 

$696  oo 

Va  44O.OC 

485.00 

607.00 

948.OO 

Vt  $614  oo 

$7l6    'CO 

$0  "?  ^    OO 

$164.4.  oo 

2 

i  cy.  vert. 

VD  $174.00 

Va  44O.OO 

$231.50 

485  .  oo 

$348  .  oo 

607.00 

$696.00 
948.00 

Vt  $614  oo 

$716  50 

$o  ^  <  .  oo 

$1644.  oo 

3 

i  cy.  horiz. 

VD  $174.00 

Va  394-50 

$231.50 

437-00 

$248.00 
540.00 

$698.00 

850  .  oo 

Vt                 $568  50 

$668  50 

$788  oo 

$1546  oo 

i  cy  horiz 

VD  ...         $174  oo 

$231  <?o 

$248.00 

$696  oo 

Va  367.00 

405.00 

500.00 

794.00 

Vt  $541  oo 

$6^6    "CO 

$748  oo 

$14.00  oo 

i  cy  vert 

VD  $130  50 

$184.  <to 

$261  .00 

$522  oo 

Va  367.00 

405.00 

500.00 

794-00 

Vt.  .             .  .  $407    =CO 

$ej8o  <;o 

$761  .00 

$1316  oo 

6 

i  cy  horiz 

VD  .  .            $161  oo 

$215  oo 

$322  oo 

$644  oo 

Va-  •                         332    OO 

367  oo 

4C2    oo 

72  3    OO 

V  t-  •                     $4Q3    OO 

$582  oo 

$77^   OO 

$1367  oo 

6 

i  cy  vert 

V  D  •  •  .         $i  16  oo 

$I<4    <?O 

$232  oo 

$464  oo 

Va  332.00 

367.00 

453-00 

723.00 

8 

i  cy.  horiz 

Vt  $448.00 

V  D    ...      .       $ISI    OO 

$521.50 
$2O2    50 

$685.00 
$302  oo 

$Il87.OO 

$604  oo 

V  a                            312    OO 

74.2    OO 

4.2  1    OO 

679  oo 

Vt-  .          .  .  $4.63  oo 

$^44  so 

$727  oo 

$1283  oo 

12 

i  cy.  horiz. 

YD  $130  50 

$184.50 

$26l  .OO 

$522  oo 

Va                              287    OO 

314.   OO 

280  oo 

628  oo 

Vt-  .               .  .  $417.  'JO 

$408  .  so 

$650.00 

$1150  oo 

12 

i  cy  vert 

VD                $11  6  oo 

$IS4    SO 

$232  oo 

$464  oo 

Va  287.00 

3I4.00 

389.00 

628.00 

7  1                 $4.03  oo 

$4.68  so 

$621  oo 

$1092  oo 

18 

i  cy.  horiz. 

VD  $116  oo 

$1^4.  <o 

$232  oo 

$464  oo 

Va  265.50 

289.50 

360.00 

584.00 

7t.  .           .  .  $381  .  <?o 

$444  .  oo 

$^Q2  .OO 

$1048.00 

148 


VESTANCES 
TABLE  71  —  Continued 


SIZE, 

LOADS 

H.P. 

100% 

75% 

50% 

25% 

18 

i  cy  horiz. 

VD   $IO2.  50 

$130  oo 

$205  oo 

$410  oo 

F24.Q    OO 

272    OO 

336  oo 

CAQ     OO 

Vt-  .          .  .  $3^1  •  ^o 

$41  i  oo 

$^4.1    OO 

$QtQ     OO 

2S 

2  cy.  vert. 

V  D    $IO2.5O 

$i  39  .  oo 

$205  oo 

$410  oo 

Ffl  259.00 

282.00 

349-00 

571.00 

Vt                 $361   <o 

$421  oo 

S  S  ZA.   OO 

$981  oo 

2C 

i  cy.  vert. 

VD  $83  .  50 

$111  .00 

$167  oo 

$334.  OO 

Fa  249.00 

272.00 

336.00 

549-00 

•TO 

i  cy  horiz. 

vt  $332.5° 

VD  $100.00 

$383.00 

$134.00 

$503  •  oo 
$200  oo 

$883.00 
$400  oo 

Fa  239.00 

260.00 

323.00 

529.00 

Vt  $339  -oo 

$394  oo 

$^23    OO 

$Q2Q    OO 

a6 

2  cy  horiz. 

VD     •  •          $116  oo 

$i  ^4  t;o 

$232  oo 

$4.64.  oo 

ou 

Fa  245.00 

265.50 

330.50 

544  .  oo 

Vt           ...  $361  oo 

$420  oo 

$^62   ^o 

$1008  oo 

36 

3  cy.  vert. 

VD  $106.00 

Fa  250.00 

$141.50 

271.50 

$212  .OO 
337-50 

$424  .  oo 
555-oo 

Vt                 $356  oo 

$4.1  ^    OO 

$^4.Q    1O 

So7o  oo 

40 

i  cy.  horiz. 

VD  $95-50 

Fa  233.50 

$127.00 
250.50 

$191.00 
312.00 

$382.00 
513-00 

Vt  $329.00 

$•277    e;o 

S^O'?  oo 

$80  c;  oo 

40 

i  cy.  horiz. 

VD  $97.50 
Fa  223.50 

$130.00 
242.50 

$195.00 

302  .  oo 

$390  .  oo 
495-00 

50 

2  cy.  horiz. 

Vt  $321.00 
VD  $97-50 

Fa  231.50 

$372.50 

$130.00 
251.00 

$497.00 

$195.00 
322.00 

$885.00 
$390  •  oo 

514.00 

Vt  $329.00 

$38l.OO 

$517  .00 

$904  .  oo 

SO 

i  cy  vert 

Fn                      $07    ^O 

$130  oo 

$IQ£f     OO 

$300  oo 

Fa  223.50 

242  .  50 

302.00 

495-oo 

ro 

2  cy  vert 

Vt  $321.00 
VD                 $83  50 

$372.50 
$111    OO 

$497-00 
$167  oo 

$885.00 
$334  .  oo 

Fa  231.50 

251.00 

322.00 

514.00 

Vt                 $3T5  oo 

$362  oo 

$4.80  oo 

$848  .  oo 

64 

2  cy.  horiz. 

VD  $97-50 

Fa  223.50 

$130.50 
242.50 

$195-50 

301  .  oo 

$391.00 
498.00 

Vt                         $3  2  1    OO 

$  2  7  2      OO 

$4.06  ^o 

$889  oo 

OIL   ENGINES 

TABLE  71  —  Concluded 


149 


SIZE, 

LOADS 

H.P. 

TYPE 

100% 

75% 

50% 

25% 

7C 

i  cy  vert 

V  n                            $OO    SO 

$13^?    OO 

$IOQ    OO 

$-208   OO 

Fa  210.50 

228.00 

284.00 

469.00 

75 

3  cy.  vert. 

Ft  $3IO-oo 
F/>  $84.00 

Fa  223.50 

$361  .00 

$112.  OO 
242.OO 

$483  .  oo 

$168.00 
301.00 

$867.00 

$336.00 
499.00 

V  t     $307  •  50 

$3=14   OO 

$469  .  oo 

$83<   00 

80 

2  cv.  horiz. 

VD  $93-5° 

Fa  217.00 

$134.00 
23I-50 

$186.50 
292.50 

$373-00 
482.00 

80 

2  cy.  horiz. 

Ft  $310.50 
VD  $96-50 

Fa  211  .OO 

$365-50 

$128.50 
224.OO 

$479-oo 

$193-00 
283.50 

$855-00 

$386.00 
470.00 

V  t              •  .  $307  <co 

$2^2  «;o 

$476  "?o 

$856  oo 

100 

i  cy.  vert. 

VD  $96  .  50 

Fa  203.00 

$128.50 
215.00 

$193.00 
273.00 

$386  .  oo 
453-00 

V  t.  .          .  .  $209.  <;o 

$343    <CO 

$466.00 

$830  oo 

IOO 

2  cy.  vert. 

VD  $96.50 
Fa  211.00 

$128.50 
224.OO 

$193.00 

283.50 

$386.00 

470.00 

IOO 

A  cv  vert 

Vt  $307-50 
VD                  $83  50 

$352.50 
$1  I  I    OO 

$476.50 

$167  oo 

$856.00 

$-2^24.   OO 

Fa  219.00 

232.00 

294.50 

490.00 

Vt  $302  50 

$343    OO 

$461    c;o 

$824  oo 

VESTANCES 


$1700 


1500 


1300 


51100 
I 

1  900 


.4.1 


50 
H.P. 


60 


70 


80 


90 


100 


FIG.  26. — Total  vestances  per  h.p.  of  horizontal  oil  engines. 


An  inspection  of  the  curve  of  total  vestance  at  full  load 
shows  that  the  prices  of  the  36  h.p.  engines  are  considerably 
too  high,  while  that  of  the  2  cylinder  25  h.p.  engine  is  also 
high.  With  these  exceptions,  the  prices  are  remarkably  con- 
sistent, with  the  prices  of  the  vertical  engines  running  con- 
tinually below  those  of  the  horizontal  engines.  Down  to  30 
h.p.  the  increase  in  total  vestance  is  not  great,  showing  good 
adaptability  for  small  isolated  plants. 

103.  Diesel  Engines.  —  For  comparison  with  other  prime 
movers  we  give  below  the  vestances  of  Diesel  engines  based 
on  a  normal  year  of  3000  hours,  interest  at  5  %,  fuel  oil  at  0.05 
cent  per  h.p.h.,  and  life  of  the  engines  15  years.  Attend- 
ance costs  are  included. 


DIESEL  ENGINES 


TABLE  72 

VESTANCES  PER  H.p.  OF  DIESEL  ENGINES 
4  CYCLE 


SIZE, 

LOADS 

H.P. 

100% 

75% 

50% 

25% 

60 

VD 

.     ..     .                  .$148.50 

$198.00 

$297.00 

$594.OO 

Va.  ... 

I37-SO 

149.00 

175.00 

246.00 

Vt 

$286.00 

$347  .00 

$472.OO 

$840  .  oo 

GO 

VD 

$120    SO 

$173  oo 

$2^0    OO 

$518  oo 

Va.-.. 

129.50 

138.00 

164.00 

224.00 

Vt 

$2  CO    OO 

$311  oo 

$423  oo 

$742   OO 

VD  .  .  . 

$123  oo 

$164.00 

$246.00 

$49  2  .  oo 

Va.  ... 

I26.OO 

136.00 

158.00 

222.00 

vt  

$249  oo 

$300.00 

$404  .  oo 

$7I4.OO 

VD 

$116  oo 

$It;c   oo 

$232  oo 

$4.64  oo 

*5° 

Va  

121  .OO 

130.00 

152.00 

209.00 

Vt 

$237    OO 

$285  oo 

$384.  oo 

$673  oo 

VD  . 

.  $  I  I  I  .  OO 

$148  oo 

$222  .OO 

$444  oo 

Va  

i  i  7  .  oo 

125.00 

146.00 

202.00 

Vt 

.  .  .                 $228  oo 

$273  oo 

$368  oo 

$646  oo 

VD  ... 

$106.00 

$141  .00 

$212.00 

$424.00 

3°° 

Va  

113.00 

120.00 

140.00 

193.00 

vt.  . 

$219  .00 

$26l  .OO 

$352.00 

$617.00 

VD 

$00    ^O 

$132  oo 

$100   OO 

$308.00 

500 

Va  

107.50 

i  i  5  .  oo 

134.00 

185.00 

vt  

$207.00 

$247.00 

$333.00 

$583.00 

>jgf\ 

VD     .. 

.  .    .           $96  oo 

$128.00 

$192.00 

$384  .  oo 

75° 

Va  

105.00 

III.  00 

130.00 

179.00 

vt 

$2OI    OO 

$239.00 

$322.00 

$563.00 

VD  ... 

.    $03.OO 

$124.00 

$186.00 

$372.00 

Va  

103.00 

108.00 

127.00 

175.00 

Vt  

$196.00 

$232.00 

$313.00 

$547.00 

152 
$900 


VESTANCES 


80( 


.700 

C-, 

EC 

1 600 


500 


100 


200 


300 


400 


700         800 


900     1000 


500         600 
H.  P. 
FIG.  27. —  Total  vestances  per  h.p.  of  Diesel  engines  at  fractional  loads. 

104.  Induction  Motors.  —  Since  the  efficiency  of  induction 
motors  is  independent  of  the  speed,  the  only  difference  in  the 
vestance  of  induction  motors  of  the  same  size  and  type  will 
be  in  the  depreciation  vestance.  Since  the  depreciation  vest- 
ance is  but  a  small  part  of  the  total  vestance,  there  will  be 
but  little  difference  in  the  cost  of  service  even  though  the 
first  cost  of  low  speed  motors  may  appear  to  be  very  large  in 
comparison  with  the  first  cost  of  high  speed  motors.  We 
illustrate  this  below. 

Example  31.  — Compare  the  total  vestance  of  a  10  h.p. 
1800  r.p.m.  3  phase  60  cycle  220  volt  induction  motor  with 
another  motor  of  the  same  size  and  type  but  of  only  600  r.p.m. 
speed.  Assume  the  normal  year  at  3000  hours,  interest  5%, 
power  cost  one  cent  per  h.p.h.  and  neglect  other  operating  costs 
besides  those  of  power. 

Solution:  The  first  cost  of  the  1800  r.p.m.  motor  is  $160 
or  $16  per  h.p.  With  25  years  of  life  and  interest  at  5%,  the 
term  factor  is  0.70469,  so  that  the  depreciation  vestance  per 
h.p.  is 

16  -5-  0.70469  =  $22.75. 


INDUCTION  MOTORS  153 

The  efficiency  of  this  sized  motor  is  87.9%  so  that  the 
power  used  per  h.p.  year  is 

3000  -T-  0.879  =  3420  h.p.h., 
costing  3420  X  o.oi  =  $34.20. 

The  operating  vestance  is  then 

34.20  -r-  0.05  =  $684, 
and  the  total  vestance  is 

22.75  +  684  =  $706.75. 

For  the  600  r.p.m.  motor  the  first  cost  is  $300  or  $30  per 
h.p.  so  that  the  depreciation  vestance  per  h.p.  is 
30  -T-  .70469  =  $42.50, 

while  the  operating  vestance  is  the  same  as  for  the  1800  r.p.m. 
motor  or  $684. 

The  total  vestance  is  then 

42.50  +  684  =  $726.50. 
The  difference  is 

42.50  -  22.75  =  $19-75 
out  of  a  total  of  $706.65  or 

19.75  -*•  7°6-75  =  2.66%. 

The  only  object  in  using  low  speed  motors  is  for  the  pur- 
pose of  combining  them  with  other  low  speed  apparatus  and 
where  this  cost  of  combination  may  be  reduced  by  an  amount 
equal  to  or  greater  than  the  additional  cost  of  the  lower  speed 
motor.  We  have  chosen  above  an  extreme  difference  in  speed. 
Where  the  difference  in  speed  is  not  so  great,  the  difference  in 
total  vestance  will  be  even  less  than  this  small  amount  of 
2.66%. 

105.  We  give  below  the  vestances  of  three  phase,  60  cycle, 
1800  r.p.m.  induction  motors  based  on  a  normal  year  of 
3000  hours,  power  cost  one  cent  per  h.p.h.  and  interest  at  5  %. 


VESTANCES 


TABLE   73 

VESTANCES  PER  H.p.  OF  INDUCTION  MOTORS 

3  PHASE,  60   CYCLES,    l8oO   R.P.M.    AT   FULL   LOAD 


SIZE 
H.P. 

VESTANCES 

Ratio 

Va 

VD 

Depreciation 

Operating 

Total 

I 

$56.80 

$767.00 

$823  .  80 

13-5 

2 

35-55 

728.00 

763.55 

20.5 

3 

26.95 

713.00 

739-95 

26.5 

5 

19.80 

700.00 

719.80 

35-4 

7-5 

25.60 

689.00 

714.60 

26.9 

10 

22.75 

684.00 

706.75 

30.0 

15 

18.20 

677.00 

695  .  20 

37-2 

20 

16.  70 

674.00 

690  .  70 

40-3 

25 

14.50 

671  .OO 

685  .  50 

46.3 

30 

14-45 

669.00 

683.45 

46-3 

40 

12.43 

667.00 

67Q-43 

53-6 

SO 

n-37 

665  .  oo 

676-37 

58-5 

75 

9.76 

663  .  oo 

672.76 

68.0 

Note  the  very  great  ratio  of  operating  to  depreciation 
vestance,  and  the  very  great  total  vestance.  This  we  shall 
discuss  further. 

Two  phase  motors  usually  run  one  to  two  per  cent  lower 
in  efficiency,  increasing  the  total  vestance  between  $7  and 
$14  per  h.p.,  a  very  appreciable  amount.  The  operating 
vestance  is  very  large  though  the  power  rate  assumed  is  far 
lower  than  that  which  can  ordinarily  be  obtained.  The  ten- 
horse  power  rate  is  usually  2  cents  per  h.p.h.  On  this  basis 
the  operating  vestance  would  be  doubled,  increasing  the  ratio 
to  60. 

106.  At  fractional  loads,  the  splendid  full  load  efficiency  of 
induction  motors  is  well  maintained,  so  that  there  is  no  great 
variation  of  vestances  at  fractional  loads.  This  is  shown  in 
the  table  below. 


INDUCTION  MOTORS 


155 


TABLE  74 
VESTANCES  PER  H.P.  OF  INDUCTION  MOTORS 

3  PHASE.   60   CYCLES,    l8oO   R.P.M. 


SUE, 

LOAD 

I 

H.P. 

125% 

100% 

75% 

50% 

25% 

V  D     •    '          .    .  .   $45  .40 

$56.80 

$75.75 

$113.60 

$227   20 

Va  ...-     7QT-00 

767.00 

767.00 

791-00 

852.50 

Vt       $836.40 

$823  .  80 

$842  .  75 

$904  .  60 

$IO7Q    7O 

2 

VD  $28.40 

$35-55 

$47  •  30 

$71  .IO 

$142.  2O 

Va  750-00 

728.00 

728.00 

750.00 

809.00 

Vt     $778.40 

$763.55 

$775-3° 

$821.10 

$051    2O 

VD           -                $25  50 

$°6  <K 

$42  50 

$53  9O 

$107   80 

Va  736-00 

713.00 

713.00 

736.00 

792.00 

Vt                        $761  50 

$7-20   or 

$75  5    5O 

$780  oo 

$800   80 

VD  $15.85 

$19.80 

$26  .  40 

$39  .  60 

$70   2O 

Va  722.00 

700.00 

700.00 

722.00 

778.00 

vt                  $737.85 

$7IQ   80 

$726.40 

$761.60 

$8^7    2O 

1\ 

VD  $20.50 

Va  711.00 

$25.60 
689.00 

$34-io 
689.00 

$51.20 
711  .00 

$102.40 
766.00 

IO 

Vt  $731-50 
VD                           $18  ^o 

$714.60 
$22    75 

$723.10 
$30  30 

$762.20 
$45    5O 

$868.40 

$91  oo 

Va  *.  .     705-00 

684.00 

684.00 

705.00 

760.00 

Vt  $723  •  20 

$706.75 

$714.30 

$750.50 

$851.00 

T  C 

VD                                $14   57 

$18  20 

$24    *O 

$36  .  40 

$72  80 

Va  697.50 

677.00 

677.00 

697.50 

752.00 

Vt                          $712  07 

$6o5    2O 

$7OI    30 

$772     QO 

$824  80 

2O 

VD  $13-37 

Va  695.00 

$16.70 
674.00 

$22.25 
674.00 

$33-40 
695.00 

$66.80 

749-oo 

Vt  $708.37 

$690  .  70 

$696.  25 

$728.40 

$815  80 

VD                            $n  60 

$14    5O 

$IO    3O 

$29  oo 

$58  oo 

25 

Va  692.00 

671  .OO 

671  .OO 

692.00 

746.00 

Vt                       •  $703  60 

$68^  =;o 

$6QO    3O 

$721  oo 

$804  oo 

30 

VD  $11-55 

Va                                          690   OO 

$14-45 
669  oo 

$I9-27 

669  oo 

$28.90 

690  oo 

$57.80 

JAA     OO 

Vt  $701  .55 

$683  45 

$688.27 

$718.00 

$801  80 

VESTANCES 
TABLE  74— Continued. 


SIZE, 

L.OAE 

s 

H.P. 

125% 

100% 

75% 

50% 

25% 

4.O 

VD                              $9  98 

$12   4.3 

$16  60 

$24  86 

Va  688.00 

667.00 

667.00 

688.00 

741.00 

Vt.  .                        $697.98 

$670   4.3 

$683  60 

$Mi2  86 

(TQ 

VD                              $9  10 

$1  1    37 

$i  ^  i  ^ 

$22    7  A 

«8>/yu.  /^ 

HUc    18 

Va  685.00 

665  .  oo 

665  .  oo 

685  .  oo 

NP45  •  4° 
739-oo 

Vt                          $694  10 

$676    37 

$680  ic 

$7O7    7A 

$78,,    AQ 

75 

VD  $7-80 

Va                                            684    OO 

$9.76 
66^    OO 

$13-00 
66^  oo 

$I9-S2 

684  oo 

sP/04-40 

$39-04 

/v5/  •  ou 

Vt..-  $691.80 

$672.76 

$676.00 

$703.52 

$776.54 

80 


FIG.  28. — Total  vestances  per  h.p.  of  induction  motors  at  fractional  loads. 

There  is  comparatively  little  change  of  total  vestance  at 
fractional  loads  because  the  efficiency  is  so  well  maintained. 
This  is  a  very  valuable  feature  for  maintaining  good  economy 
under  actual  load  conditions,  such  as  are  invariably  met  in 
practice. 

107.  Generators.  —  The  vestances  for  alternating  current 
generators  of  the  direct  connected  type  are  given  below  based 


GENERATORS 


157 


on  a  normal  year  of  3000  hours,  cost  of  energy  £  cent  per  kv.a., 
interest  5%  and  a  life  of   25  years. 

TABLE  75 
VESTANCES  PER  KV.A.  OF  D-c.  GENERATORS 

3   PHASE,   60   CYCLES,    220-550   VOLTS 


SIZE, 

LOADS 

KV.A. 

125% 

100% 

75% 

50% 

25% 

IO 

VD  $28.40 

Va  208.10 

$35-50 
205.00 

$47-30 
209  .  80 

$71.00 
222.80 

$142.00 
257.00 

Vt              .       .     $236.50 

$240.  so 

$257.  10 

$2(H    80 

$300   OO 

2< 

VD  $22.70 

$28.40 

$37  -  80 

$56  .  80 

$113.60 

Va  182.00 

178.00 

182.50 

193.00 

217.00 

Vt                          $204  70 

$206  40 

$220  30 

$249    80 

$•230    60 

VD  $18.80 

$23  .  <?o 

$31  .  30 

$47  .00 

$04  OO 

5° 

Va  172.00 

169.00 

172.50 

184.00 

205.00 

Vt  $190.80 

$192.50 

$203.80 

$231  .00 

$200    OO 

VD                             $17  oo 

$21    30 

$28  « 

$42  60 

$85    2O 

75 

Va  169.50 

166.80 

170.25 

181  .00 

202.50 

Vt                            $186  50 

$188.10 

$198  60 

$223  60 

$287    7O 

V  D    $15  -9O 

$IQ.QO 

$26.50 

$19.  80 

$79  60 

IOO 

Va  167.25 

154.10 

•168.00 

178.75 

200.00 

Vt.  .                       .$183.15 

$184.00 

$104.  50 

$2l8.<X 

$270    60 

150 

VD  $H.8o 

Va  165.40 

$18.50 
162.75 

$24.70 
166.75 

$37.00 
177.50 

$74-oo 
197-50 

Vt                            $180  20 

$181   25 

$IOI    A< 

$214.    SO 

$271    <?O 

250 

VD  $13.00 

Va  164.00 

$16.30 
161  .25 

$21.70 
165.00 

$32.60 
175.00 

$65  .  20 
I95.50 

Vt-  •  .  .                    $177  .00 

$177    "?^ 

$186.70 

$207.60 

$260    70 

VD                              $n   70 

SlA    2O 

$18  90 

$28  40 

$<c6  80 

500 

Va  162.25 

I59.50 

162.75 

173-25 

193-25 

Vt-  .                                 $173   0^ 

$173   70 

$181  65 

$201  .65 

$2=;o  os 

750 

VD  $10.85 

Va-  •  •                                      l6l    OO 

$I3-65 
i  ^8  c,o 

$18.20 
162  25 

$27.30 

172    £,O 

$54  .  60 

IQ2    SO 

Vt  $171  85 

$172"  is 

$180  45 

$100  -80 

$247  .  10 

VD                              $10  67 

$i  3  ^s 

$17  80 

$26    70 

$S'l    4.O 

Va  1  60  50 

I  C.8   oO 

161  7"\ 

1  72  oo 

IQI    7"> 

i 

Vt                           $171  17 

$171           3S 

Sl7O    S  S 

$198    70 

$24.  e,    I1? 

158 


VESTANCES 


$1200 
1100 


1000 


900 


700 


600 


700         800        900      1000 
FIG.  29. — Total  vestances  per  kv.a.  of  electric  generators  at  fractional  loads. 


0  100         200         300         400         500         600 

"K.  V.  A. 


Like  motors,  there  is  comparatively  little  variation  in  total 
vestance  with  the  size  of  the  unit  or  the  load.  This  is  one  of 
the  immensely  valuable  characteristics  of  electrical  machinery 
that  it  is  so  well  adapted  to  meet  the  load  variations  of  prac- 
tice with  remarkably  good  economy. 

108.  Comparison  of  Power  Units.  —  By  power  units  we 
mean,  primarily,  such  units  as  steam  engines,  turbines,  inter- 
nal combustion  motors,  and  electric  motors.  In  the  fore- 
going, we  have  determined  the  vestances  of  these  units  based 
on  a  normal  year  of  3000  hours,  constant  load,  and  5  %  interest. 
The  fuel  costs  assumed  were  coal  at  $2  per  ton  with  a  fuel 
value  of  13,743  B.t.u.  or  5.4  h.p.h.  per  Ib.  Allowing  for  boiler 
service,  we  assumed  a  cost  of  $0.005  Per  h.p.h.  of  energy  de- 
livered to  the  engine.  In  comparison  with  this  we  assumed 
for  oil  engines,  oil  at  0.511  cent  per  pound,  and  allowing  320 
Ibs.  per  barrel,  this  would  amount  to  $1.635  Per  barrel,  or 
0.0^7  cent  per  h.p.h.  For  Diesel  engines  we  assumed  the 
cost  of  the  heat  energy  at  0.05  cent  per  h.p.h.  corresponding 
to  a  cost  of  $1.168  per  barrel.  And  finally  for  motors  we 
assumed  a  cost  of  one  cent  per  h.p.h.  At  15  cents  per  gallon 


COMPARISON  OF  POWER  UNITS  159 

for  gasoline,  energy  would  cost  0.2675  cent  per  h.p.h.     For 
clearness  we  tabulate  these  as  follows: 


TABLE   76 

COST  OF  ENERGY  PER  H.P.H.  ASSUMED 

COST  OF  ENERGY  PER 
UNIT  H.P.H.  IN  CENTS 

Induction  motor i .  oooo 

Gasoline  engine o .  2675 

Oil  engine 0.0700 

Diesel  engine 0.05 

Steam  engine 0.05 

With  these  assumed  costs  clearly  in  view,  for  they  vary 
greatly  with  different  localities,  we  found  comparisons  as 
follows: 

At  full  load,  the  Diesel  engine  has  the  best  economy,  with 
the  oil  engine  second,  the  compound  condensing  steam  engine 
third,  the  simple  low  speed  noncondensing  steam  engine 
fourth,  the  simple  high  speed  noncondensing  steam  engine 
fifth,  the  electric  motor  sixth  and  the  gasolene  engine  seventh. 

Below  30  h.p.  the  electric  motor  is  far  superior  in  economy 
to  the  simple  high  speed  steam  engine,  and  no  doubt  the  other 
types  of  steam  engines  as  well,  but  it  is  itself  far  inferior  to  the 
oil  engine,  somewhat  inferior  to  the  distillate  engine  at  7^ 
cents  per  gallon  for  the  distillate,  and  far  superior  to  the  gaso- 
line engine.  The  sharp  bending  of  the  curve  (Figure  30)  of 
the  Diesel  to  the  horizontal  for  small  sizes  would  lead  one  to 
expect  the  oil  engine  to  be  superior  to  the  Diesel  below 
40  h.p. 

109.  We  give  below  in  tabular  form  the  total  vestances  of 
these  units  for  both  full  and  fractional  loads,  together  with 
the  relative  valuance  based  on  unity  for  the  machine  which 
has  the  best  economy,  so  as  to  show  clearly  their  compara- 
tive values.  Where  the  values  are  not  given  in  the  tables 
they  have  been  interpolated  from  the  curves.  In  the  table, 
Vt  is  the  total  vestance  and  U,  the  comparative  value. 


160 


VESTANCES 


TABLE   77 
TOTAL  VESTANCES  AND  COMPARATIVE  VALUE  OF  POWER  UNITS 


SIZE, 
H.P. 

UNIT 

LOADS  PEK  CENT  OF  FULL  LOADS 

125 

IOO 

75 

50 

25 

Vt 

U 

Vt 

V 

Vt 

U 

V 

U 

0.98 

I  .00 

0.80 

0-75 

0.99 
0.90 
i  .00 
0.83 
0.77 

0.98 

I  .00 

0.82 

0.70 

1  .00 
0.42 
0.66 
o.  70 

Vt 

U 

0.7 

I  .OC 

0.8. 
0.8 

o.7< 

o.6( 

I  .OC 

0.9 

0.8 

o.8c 
o.8v 

I.OC 

0-9, 

0.91 

0.3^ 
o.8( 

I  .OC 

250 
IOO 

50 
25 

Diesel  e 
Comp.  < 
S.  L.  S. 
S.  H.S. 

Diesel  e 
Oil  engi 
Comp.  s 
S.  L.  S. 
S.H.  S. 

Diesel  t 
Oil  engi 
S.  H.  S. 
Electric 

Oil  engi 
Gasoline 
S.  H.  S. 
Electric 

ngii 
;tea 
stes 
stej 

ngii 
ne. 

ie  

223 
300 
386 
426 

254 
300 

339 
429 

474 

300 
321 
530 
676 

332 
864 
660 
686 

i  .00 
0.74 
0.58 
0.52 

I.OO 

0.85 
o-75 
o-59 
0-54 

i  .00 
0.94 

0-5.7 
0.44 

I.OO 

0.38 
0.50 
0.48 

267 
315 
396 
434 

306 

345  < 
361  < 
443  < 

485  < 

365 
373  < 

545  < 
680  c 

383 
960  c 
682  c 
690  c 

i  .00 
3.85 
3.67 

D.62 

[  .00 

3.89 

3.85 
3.69 

3.63 

[.00 

3.98 

1.67 

3-54 
c  .00 

5.40 

3.56 
5.55 

36c 

44c 
474 

4M 

41: 
49^ 
532 

497 
603 

503 

1212 

757 
721 

631 

•457 
543 
560 

728 
839 
544 
609 

637 

910 
885 
732 
785 

883 
2089 
936 
804 

m  engine.  .  .  . 
im  engine  .  .  . 
im  engine.  .  . 

ie.  .  . 

307 
4.02 

448 

I.OO 

0.76 
0.69 

teai 

StCc 

ste« 

-ngi 
ne. 

"n  engine.  .  .  . 
im  engine.  .  . 
im  engine.  .  . 

ne  

345 
14  2 

199 

I.OO 
D.78 

3.69 

ste 
mo 

ie. 

am  engine  .  .  . 
tor  

556 
594 

[.OO 

3.80 

!  enj 
ste£ 
mo 

Tine 

un  engine.  . 
tor 

>95 
ro4 

[  .00 

3.99 

G 

—  - 

solin 

-—  . 

(Em 

fiflO 

700  V 

W 

X  _\ 

Klec 

trie  Motoq 

a  600U— 

\ 

I 

S 

\. 

1  J\ 

^^ 

V 

400h- 

\ 

1  

•  —  - 

•  —  . 
~  i— 

-  — 
—  — 

— 

••    —. 

•-•a^*i 
—     i— 

—  ^^ 
•        •• 

S 

1 

r*  •• 

MS. 

-  —— 

5.L.S. 

•  •  i  •• 

^on  t 

-       • 

ion  < 

mmmtfst 

mde 
nde 

jiin 
_sinj 

•' 

__ 

300H 

^. 

•Ov 

Oil  F 

npin 

-  — 

—  — 
—  — 

••     •• 

•          ••• 

= 

C 

)iese 

trip 

_ET 

5  nd  ] 

•  • 

S.  S 

—  1    •• 

iam 

•  

r-«-i 

^ 

*^ 

•  

•  1 

-— 

-t- 

ii       •• 

*°°0            25           50           75          100         125         JL50         1.75         200         225         25C 
H.  P. 

FIG.  30.  —  Total  vestances  of  power  units  at  full  load. 


COMPARISON  OF  POWER  UNITS  161 

It  is  evident  from  the  above  table  that  at  full  load  the 
Diesel  engine  leads  in  economy  by  a  wide  margin,  with  the 
oil  engine  second  and  the  compound  steam  engine  third.  For 
sizes  over  50  h.p.,  the  oil  engine  is  worth  only  85%  as  much 
as  a  Diesel  and  the  compound  steam  engine  about  75%.  In 
other  words,  taking  the  cost  of  service  of  the  Diesel  as  unity, 
then  that  of  the  oil  engine  is  1.15,  i.e.,  15%,  greater  and  the 
cost  of  service  with  compound  condensing  engines  is  1.33  or 
33%  greater. 

At  75%  load  the  Diesel  still  leads  but  at  50  h.p.  the  oil 
engine  has  practically  become  equal  to  it.  Diesel  engines 
are  seldom  built  below  this  size,  though  a  few  have  been  made 
as  small  as  8  h.p.  It  is  certainly  remarkable  how  well  the 
designers  of  engines  have  chosen  the  inferior  limit  of  the 
sizes  of  their  various  types. 

At  50%  load,  the  Diesel  and  compound  engines  have 
reached  equality  in  sizes  of  100  h.p.  and  over,  with  the  oil 
engine  third.  But  below  50  h.p.  the  oil  engine  is  best. 

At  25%  load,  the  compound  steam  engine  leads  by  a  wide 
margin  above  100  h.p.  At  50  h.p.  the  simple  high  speed  steam 
engine  is  best,  with  the  electric  motor  second,  while  below  25 
h.p.  the  electric  motor  leads  with  the  oil  engine  second,  based, 
however,  on  costs  of  energy  as  given. 

In  all  this  the  steam  engines  have  the  peculiar  advantage 
that  they  can  carry  with  comparative  good  economy  25% 
overload.  The  range  of  loads  of  these  engines  is  therefore 
25%  greater  than  internal  combustion  motors.  The  advan- 
tages of  this  greater  range  of  load  will  be  shown  later. 

The  gasoline  engine  is  inferior  to  the  electric  motor  at  all 
loads  on  the  basis  chosen.  But  if,  for  example,  2  cents  per 
h.p.  is  paid  for  electric  power  as  compared  to  15  cents  per 
gallon  for  gasoline,  then  conditions  would  be  reversed. 

In  the  face  of  the  great  superiority  of  the  Diesel  engines,  it 
seems  remarkable  that  they  are  not  more  generally  used. 
There  are  two  reasons  for  this.  First  the  general  resistance 
of  engineers  to  anything  new  to  them,  and  second  the  fact 
that  there  are  many  cases  where  the  undertakings  are  under- 


l62 


VESTANCES 


financed,  so  that  they  are  unable  to  put  in  the  most  economical 
installations.  Above  all  this  looms  the  element  of  uncer- 
tainty, so  that  there  is  a  lack  of  incentive  to  build  for 
permanency. 

110.  Centrifugal  Pumps.  —  In  the  case  of  centrifugal 
pumps,  the  costs  of  attendance,  oil,  and  the  like,  are  very 
small,  as  compared  with  the  cost  of  power.  In  the  following 
tables  of  vestances  of  centrifugal  pumps,  we  have,  therefore, 
left  these  items  out  of  consideration.  The  vestances  are  given 
not  only  at  full  and  fractional  loads,  but  at  variable  heads  as 
well.  We  have  assumed  a  cost  of  power  at  one  cent  delivered 
at  the  pump  shaft,  interest  rate  5%  as  before,  and  a  normal 
year  of  operation  at  1500  hours,  with  25  years  as  the  life  of 
the  pumps. 

TABLE   78 

VESTANCES  PER  1000  G.P.M.  OF  CAPACITY  OF  Low  HEAD  BELTED  CENTRIFUGAL 

PUMPS 


RATED 

DEPRECIA- 
TION VES- 

OPERATING  VE 

STANCES  AT  HE, 

\DS   OF 

CAPACITY  , 

G.P.M. 

TANCE  PER 

IOOO   G.P.M. 

10' 

20' 

30' 

50' 

50 

$936.60 

$3400  .  oo 

$6000  .  oo 

$7500.00 

$11,360.00 

100 

596.10 

268O.OO 

4410.00 

592O.OO 

9,360.00 

150 

482.50 

2350.00 

4050  .  oo 

5370.00 

8,320.00 

225 

377-50 

2O25  .OO 

3625.00 

4670.00 

7,430.00 

300 

326.40 

1875.00 

3330.00 

4420  .  oo 

7,070.00 

400 

265.40 

i  790  .  oo 

3120.00 

4170.00 

6,690.00 

700 

2O2  .  80 

1565.00 

2770.00 

3750.00 

6,040.00 

900 

189.  2O 

1445  .  oo 

2540.00 

3570.00 

5,850.00 

I,2OO 

177.20 

1390.00 

2500.00 

3460  .  oo 

5,680.00 

1,  600 

160.00 

1320.00 

2380  .  oo 

3360  .  oo 

5,500.00 

3,000 

119.50 

1295.00 

2320.00 

3310.00 

5,460.00 

4,5°o 

106.40 

1270.00 

2270.00 

3260.00 

5,400.00 

6,000 

96.70 

1250.00 

2230.00 

3220.00 

5,340.00 

7,000 

93-40 

1230.00 

2220.00 

3170.00 

5,310.00 

8,000 

91  .  2O 

I2IO.OO 

22OO.OO 

3120.00 

5,280.00 

10,000 

87.30 

i  i  90  .  oo 

2l8o.OO 

3080  .  oo 

14  ooo 

8?  AQ 

i  185  .  oo 

2  170  .  OO 

304.0  oo 

20,000 

O  *  T" 

So.OO 

i  i  70  .  oo 

2160.00 

O    ^      ' 

3000  .  oo 



•7Q  OOO 

78.00 

I  j  5  tr    OO 

2  I  40  .  OO 

O'-'j1-"-"-' 

40,000 

76.80 

I  l6o.OO 

2135.00 

50  ooo 

76  .  2O 

1  1  ee    OO 

2130  .  oo 

60  ooo 

7^    7O 

•*•  L  o  o  •  ^^ 

i  150  .  oo 

2125  .00 

/  o  •  p* 

CENTRIFUGAL  PUMPS 


163 


Total  Vestance  ^ 

11 

\ 

"v 

__ 

I 

\ 

s^ 

9,0-ft 

-Tfri 

d  — 

V 

10-ft 

H 

d  

0         2.000       400.0      6000       8000      10000     12000    14000     16000     1SOOO   20000 
Capacity  in  G.P..M. 

FIG.  31. — Total  vestances  of  low  head  belted  horizontal  centrifugal 
pumps  at  full  load. 


$16000 


200        400        600         800        1000       1200       1400       1600       1800 
Capacity  in  G.P.M. 

FIG.  32.  —  Total  vestances  of  vertical  centrifugal  pumps. 


i64 


VESTANCES 


TABLE   79 

VESTANCES  PER  1000  G.P.M.  OF  CAPACITY  OF  VERTICAL  CENTRIFUGAL 

PUMPS 


CAPACITY, 

G.P.M. 

VESTANCES  AT  HEADS  OF 

,5' 

50' 

75' 

IOO 

ISO 
225 
300 
400 
700 
900 
1  200 
1600 

Va 

$5210  oo 

$9,370.00 

3,760.00 

$13,700.00 
4,740.00 

VD  

2780.00 

Vt 

$7QOO    OO 

$13,130.00 

$8,330  .  oo 
2,770.00 

$18,440.00 

$12,300.00 
3,440.00 

Va  

$4690  oo 

VD  

vt  

2090  .  oo 

$6780.00 

$11,100.00 

$7,500.00 

2,100.00 

$15,740.00 

$II,O5O.OO 
2,670.00 

Va 

$4150.00 
1570.00 

VD  

vt 

$s72O   OO 

$9,600.00 

$7,070.00 

1,740.00 

$13,720.00 

$IO,42O.OO 
2,I4O.OO 

Va  

$3980  .  oo 

VD  

vt  

1330.00 

$5310  oo 

$8,810.00 

$6,690.00 
i  ,405  .  oo 

$I2,56o.OO 

$9,890.00 
I,7IO.OO 

Va 

$3750.00 
1063  oo 

VD 

Vt 

$4813  oo 

$8,095  .  oo 

$6,040.00 
938  .  oo 

$II,60O.OO 

$8,930.00 
I,I4O.OO 

Va  

$3290  oo 

VD  

717.00 

vt  

$4007  oo 

$6,978.00 

$5,860.00 

866.00 

$IO,O7O.OO 
$8,650.00 

i  ,040  .  oo 

Va 

$3020.00 
670  oo 

VD 

vt 

$3690  oo 

$6,726.00 

$5,680.00 
727.00 

$9,690.00 

$8,410.00 
866.00 

Va..  .         . 

$202^    OO 

VD  

S87-00 

Vt  

$3512  oo 

$6,407.00 

$5,510.00 
657.00 

$9,276.00 

$8,160.00 
762.00 
$8,922.00 

Va 

$2880  oo 

VD  .  . 

378  oo 

Vt 

$  3  2  tj  8  oo 

$6,167.00 

CENTRIFUGAL  PUMPS 


165 


1 

8OO           OOO           000           OOO           OOO 
00           OOO           OOO           OOO           OOO 

OOO           OOO           OOO           OOO           OOO 
OOO           O    ^o     to          O   ^O     ^O           O    *o     lo          O    CN      (N 

O    c^      <N           O  OO     OC           O   \O     ^O           co   c^      to          O     ^"     ^" 

8  8  8 

8  a  a 

10  *O      M 

stances  at 
of  150' 

cocoes          OOO          ior^M          co-^-t^.         o    ^     w 
IOMI^          IOMO           ^           lo          co            co          co            co 

CO            co 

4    •      •             i  4   •      *             >  &   • 

:    ; 

Ufc^tT     ^t-u     t^^tT     ^^tT     fc».  u-  fcT 

^  ^  fck 

i 

t^-                       O                        O                        O                       to 

I—  t                            CO                          to                           ON                           M 

CO                          to                         H|N                           O                            to 

! 

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888     888     888     888     888 

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O     co     co          Oc^<N           OOO           O     M      M           O     co     co 

8  8  8 

O    co     co 

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M      ^J-      ^O              H      H        <N              OC      W        O^              1O                *O              ^t"                *O 

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Capacity 
h.p.  g.p.m. 

to                     to                     o                      to 
<O                         co                         w                          O                          M 

Tt*                                    M                                     M                                     CO                                    CO 

r                i                 i                i                i 

co                       ^-O                     HN                       to                      *o 

20-470 

888     888     888     888     888 

8    8     8 

l 

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Ototo          O'^J'Tj-          Ococo          O    lr>    ^O          Ococo 
Oto^o          OOOOO           O^OO          toO^"^         toQto 

O  00    00 

<N      CN        Tj- 

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1 



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t?^u     ^^£     ;^^£     ^sf  tT     t^u  ^T 

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Capacity 
h.p.  g.p.m. 

to                     O 

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1 

| 

888     888     888     888     888 

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Capacity 
h.p.  g.p.m. 

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VESTANCES 


1 

800         ooo         ooo         ooo 
oo        ooo        ooo        ooo 

OOO            OOO           O    10    10          OOO 
O     co     co            O     M      M             O    f°**     t^*            O   OO     OO 

OOO                 O      ^        <N                 O      CM        <N                 M      TJ-      10 

888      88 

666        6  10 

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C^\   co     O4             t^*   O 

0 
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10 

00 

t/1  JJ 

O      M        CS                 <N      M        CO               M      M        <N                 ONMQ 
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t"^.     H         Q\               OO       M 

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1 

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Capacity 
h.p.  g.p.m. 

810                        O                         O 
CO                                          O                                           !O 

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1                        1                        1                        1 
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888      888      888      888 

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h.p.  g.p.m. 

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Capacity 
h.p.  g.p.m. 

o                  o 
80                   o                   o 
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3 

888      888;      888      888 

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888      8  8 

d  oc   06         d  <N 

co   ^"     J>-            O     t"*» 
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8 

ft 

33     £     3     £     S     £     S 

1  :*     *  : 

4 

i 

•      ;       i                    ; 

^  ^  ^T      ^  ^ 

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Capacity 
h.p.  g.p.m. 

§o 
o 
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1         r          r 

H*«                            O                              0                             JJ 

o                   o 

10                        o 

co                           O 

M                                    <N 

k                         0 
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CENTRIFUGAL  PUMPS 


167 


1 

88888888888888;8888 

O    to    to    O    to    to    O    CN      oj    O     to    to    O    O     O     O    O     O 
to  O     to    O    O     O     to  Qs    ^J"  o     H      H     O    CN     CM     to  co   oo 
*O     cs     OQ      O    <N      01      t^**  10     co   HH      O\     O      O    to     to   \o    t^»     co 

P 

M                    IH         ^                   <N         M                    MM                      MCS                    CS         M                    M 

1 

:    :            ;    •    :    :            :                •        : 

^^^^^^U^UH^^tC^^^UI^I^ 

1  Capacity 
h.p.  g.p.m. 

0             S             8             2 
o              o              o              o              o              o 

10                  o                   w                   ^i-                  10                 OO 

OO                            VO                              M                              M                              M                              M 

i                  i                  10                 10                 0                  i 

10                        >0                        r»                        I"-                        O                         O 

M                              M 

• 

1 

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Capacity  in  G.P.M. 

FIG.  33. — Total  vestances  of  horizontal  direct  connected  centrifugal  pumps. 

111.  Nothing  could  show  better  how  vestance  indicates  di- 
rectly the  relative  values  of  various  equipment  Than  table  80. 
Where  there  are  so  many  different  sizes,  speeds,  capacities,  etc., 
each  with  a  different  efficiency  and  price,  it  would  seem  at 
first  that  there  are  so  many  variables  that  a  basis  of  compari- 
son could  not  be  found.  When,  however,  all  these  variables  are 
reduced  to  the  basis  of  a  permanent  rendition  of  the  service,  the 
problem  simplifies  at  once  and  we  have  the  total  vestance. 
That  unit  which,  under  given  conditions,  has  the  lowest  total 
vestance  is  the  most  economical,  i.e.,  has  the  highest  financial 
efficiency. 

Thus  at  25  ft.  head,  the  20  h.p.  2300  g.p.m.  pump  has  the 
lowest  total  vestance  of  $2919  per  1000  g.p.m.  of  capacity, 
the  next  lowest  being  the  30  h.p.  3200  g.p.m.  pump,  having 
a  total  vestance  of  $3042.  Each  of  these  pumps  has  a  first 
cost  of  $500,  apparently,  a  rather  high  price.  Yet  the  first  cost 
represents  in  this  case  only  about  one-sixth  of  the  total  cost 
of  rendering  the  service. 

In  the  case  of  the  20  h.p.  2300  g.p.m.  pump,  the  first  cost 
per  1000  g.p.m.  of  capacity  is  $218  and  the  total  depreciation 


CENTRIFUGAL  PUMPS  169 

vestance  is  $309.  The  difference  of  $91  is  due  to  limited  life. 
If  then  this  pump  could  be  so  constructed  that  its  life  would 
be  permanent,  our  total  vestance  would  be  reduced  only  by  this 
amount  of  $91,  not  a  very  significant  per  cent  of  the  total 
vestance.  It  is  evident  then  that  increasing  the  life  of 
pumps  is  not  nearly  as  important  as  increasing  the  pump 
efficiency. 

On  the  other  hand,  since  the  operating  vestance  is  $2610,  a 
saving  of  10  per  cent  of  the  power  would  reduce  our  total 
vestance  $261,  a  very  significant  amount.  The  question 
naturally  arises  as  to  how  much  the  first  cost  could  be  in- 
creased under  the  above  conditions  for  a  saving  of  ten  per 
cent  of  the  power  used.  This  is  obtained  by  multiplying  the 
saving  in  total  vestance  by  the  term  factor  which  is  0.70469 
in  this  case.  This  gives 

261  X  0.70469  =  $183.92. 

If,  then,  a  pump  of  2300  g.p.m.  requiring  20  h.p.  is  worth  $500, 
then  a  pump  of  the  same  capacity  requiring  10  per  cent  less 
power  or  18.0  h.p.  will  be  worth 

500  -f  183.92  =  $683.92. 

If,  for  example,  the  cost  of  the  latter  were  $600,  it  would  still 
save  the  customer  $83.92  and  this  is  for  each  10  per  cent  of  the 
power  saved.  Such  improvement  by  refinement  in  design  and 
workmanship  and  by  the  use  of  better  materials  is  attainable 
to  the  extent  of  some  5  %  on  the  average  pump  built. 

It  should  be  noted  that  according  to  the  above  list  at  25 
ft.  head,  it  would  be  more  economical  to  use  five  20  h.p.  2300 
g.p.m.  pumps  than  two  50  h.p.  4800  g.p.m.  pumps  on  a  100  h.p. 
installation.  In  the  former  case  the  pump  cost  will  amount  to 

5  X  500  =  $2500, 
and  in  the  latter  case  to 

2  X  675  =  $1350. 

The  saving  in  use  of  the  5-20  h.p.  pumps  in  total  vestance 
over  that  of  two  50  h.p.  pumps  as  above,  will  be 

$3324  -  2919  =  $405 


170     /  VESTANCES 

for  each  1000  g.p.m.  of  capacity  or  a  total  of  $4657.50,  cer- 
tainly a  significant  sum. 

At  50  ft.  head  the  40  h.p.  2600  g.p.m.  pump  is  best,  the 
next  best  being  the  35  h.p.  200  g.p.m.  pump,  the  difference  in 
total  vestance  being 

$5558  -  5i35  =  $423- 


At  100  ft.  head  the  250  h.p.  7000  g.p.m.  pump  is  best 
with  the  200  h.p.  5600  g.p.m.  pump  a  close  second. 

Finally  at  150  ft.  head,  the  75  h.p.  1400  g.p.m.  pump  is  best 
with  the  100  h.p.  1800  g.p.m.  pump  as  a  close  second. 

112.  It  will  be  noted  that  the  largest  sized  pump  does  not 
always  represent  the  best  investment  considering  the  way  pumps 
are  now  being  manufactured.  However,  if  it  is  possible  at  50 
ft.  head  to  make  a  40  h.p.  pump  with  a  total  vestance  of  only 
$5135  it  certainly  is  possible  for  the  manufacturer  to  make  a 
150  h.p.  pump  with  an  even  lower  total  vestance  instead  of 
$830  higher  per  1000  g.p.m.  of  capacity.  It  is  apparent  that 
the  design  of  the  latter  needs  revision.  In  the  future  when 
a  new  design  has  been  made  and  tested,  it  will  have  to  be 
valuated,  the  results  of  which  will  determine  whether  it  will  be 
adopted  or  junked. 

In  the  list  of  pumps  at  25  foot  head  are  two  10  h.p.  pumps, 
one  of  which  costs  $240  and  delivers  820  g.p.m.,  while  the 
other  costs  $385  and  delivers  1000  g.p.m.  The  difference  in 
costs  is  $145  and  the  difference  in  capacity  180  g.p.m.  Usually 
the  latter  pump  is  considered  to  be  so  expensive  that  its  use 
is  prohibitive  except  in  "  fancy"  installations.  Yet  the  differ- 
ence in  vestance  of  $520  or  nearly  15%  shows  that  in  reality 
the  use  of  the  cheap  pump  is  prohibitive. 

At  high  heads,  good  efficiency  is  still  more  important  than 
at  low  heads.  At  150  ft.  head  there  are  listed  two  100  h.p. 
pumps,  one  having  a  capacity  of  1500  g.p.m.  and  the  other  a 
capacity  of  1800  g.p.m.  The  latter  has  a  first  cost  of  $925, 
and  the  former  a  cost  of  $550,  a  difference  of  $375.  That  is 
the  latter  costs  68%  more  than  the  former.  It  is  often  said 
that  such  nne  pumps  as  the  latter  are  installed  merely  for 


CENTRIFUGAL  PUMPS  171 

show.  Yet  the  difference  in  total  vestance  in  favor  of  the 
latter  is 

20,520  -  17,380  =  $3140 
or  3140  +  20,520  =  15.3%. 

This  means  that  in  the  latter  case  the  service  can  be  rendered 
15-3%  cheaper. 

It  should  be  borne  in  mind  that  the  cost  of  power  assumed 
is  very  low,  much  lower  than  can  usually  be  obtained.  The 
effect  of  increased  power  cost  is  to  make  an  increase  of  effi- 
ciency more  valuable,  and  to  increase  the  difference  in  vestances 
of  units  whose  efficiencies  differ. 

113.  It  is  ordinarily  assumed  that  pumps  are  either  used 
at  full  capacity,  or  not  at  all.  This  is  true  only  to 
the  extent  that  pumps  are  not  throttled.  But  almost  in- 
variably the  total  head  pumped  against  varies  with  the  season, 
i.e.,  with  the  change  in  level  between  the  supply  water  and  the 
point  of  discharge.  The  point  of  discharge  is  usually  constant, 
but  the  level  of  the  supply  water  almost  invariably  changes. 
In  the  case  where  the  water  is  pumped  from  a  river,  the  water 
level  may  change  as  much  as  15  or  20  ft.  The  effect  of  this 
is  that  if  the  head  is  decreased  below  normal,  the  capacity  of 
the  pump  is  increased  above  normal,  i.e.,  the  pump  is  over- 
loaded, while  if  the  head  is  increased  above  normal,  the  pump 
will  be  underloaded.  The  amount  of  load  variation  is  very 
small  in  high  head  pumping  plants  and  usually  large  in  very 
low  head  plants. 

It  is  evident,  then,  that  the  effect  of  fractional  loading  can- 
not be  neglected  since  the  efficiency  of  pumps  falls  off  quite 
rapidly  with  either  increase  or  decrease  of  capacity,  especially 
the  former.  Unless  we  take  into  consideration  the  amount  of 
water  to  be  pumped  at  each  stage  of  water  level,  the  cost  of 
service  will  be  inexplicably  increased.  And  it  must  be  borne 
in  mind  that  a  very  few  feet  variation  in  head  will  change 
the  pump  capacity  very  greatly,  this  being  especially  true  of 
increased  heads  above  normal.  An  idea  of  the  effect  of  all 
this  may  be  obtained  from  the  following  example. 


172  VESTANCES 

Example  32.  —  A   centrifugal   pump   delivers    1000   g.p.m. 
against  66  ft.  head  at  60%  efficiency,  this  being  its  point  of 
best  efficiency.     Assuming  a  normal  irrigation  year  of  1500 
hours,  cost  of  power  at  i  cent  per  h.p.h.  and  interest  at  5%, 
and  neglecting  other  operating  costs,  determine  the  operating 
vestance  at  full,  three-quarters,  half,  and  one-fourth  loads. 
Solution:  The  power  used  at  full  load  is 
66  X    1000  _  , 

4000  X  0.60  ~~ 

costing  27^  cents  per  hour  or 

0.275  X  1500  =  $412.5  per  year. 
The  operating  vestance  is  then 

412.5    -r-   0.05    =    $8250. 

At  three-fourths,  or  one  and  one-fourth  load,  this  would  be 
8250  -T-  0.937  =  $8804.70, 

since  at  that  load  the  pump  only  gives  93.7%  of  full  load 
efficiency. 

So  at  half  or  one  and  one-half  loads,  the  operating  vestance 
is  8250  -h  0.75  =  $11,000, 

and  at  one-fourth  load  it  is 

8250  -T-  0.436  =  $18,920. 

114.  It  is  evident  that  at  less  than  three-fourths  or  over  one 
and  one-fourth  load,  the  operating  vestance  becomes  prohibi- 
tive, unless  the  pump  is  used  but  a  few  hours  per  year.  The 
effect  of  the  number  of  hours  of  use  per  year  and  the  location 
of  the  change  point  is  similar  to  that  of  steam  engines,  except 
that  in  this  case  we  have  assumed  the  normal  year  at  1 500, 
instead  of  3000,  hours. 

Thus  if  Fo  is  the  operating  vestance  of  pumps  per  normal 
year  as  listed  in  the  preceding  tables,  then  the  operating  vest- 
ance per  hour  is  — —  and  the  total  vestance  per  year  of  ( N) 
1500 

hours  is 

V=VD+™. 
1500 


CENTRIFUGAL  PUMPS  173 

Equating  the  total  vestances  of  any  two  pumps  will  deter- 
mine the  change  point. 

Example  33.  —  At  150  ft.  head,  two  75  h.p.  direct  connected 
pumps  are  listed,  one  having  a  capacity  of  1200  g.p.m.  and 
costing  $500  and  the  other  having  a  capacity  of  1400  g.p.m. 
and  costing  $900.  Determine  the  change  point  on  the  basis 
previously  assumed. 

Solution:   The  operating  vestance  of  the  1200  g.p.m.  pump 

per  hour  is 

18,750  -;-  1500  =  $12.50. 

The  total  veatance  per  year  of  (N)  hours  is 

Fi  =  592  +  12.5  Af. 

For  the  1400  g.p.m.  pump,  the  operating  vestance  per  hour 
is  16,100  -4-  1500  =  $10.733, 

so  the  total  vestance  per  year  of  (N)  hours  is 

F2  =  915  +  10.733  N. 

The  change  point  is  determined  by  equating  the  total  vest- 
ances, i.e. 

592  +  12.5  #1  =  915  +  10.733  #1, 
or  Ni  =  182.8  hrs. 

If  the  pumps  are  to  be  used  less  than  182.8  hours  per  year 
the  cheaper  pump  is  best;  for  more  than  this  number  of  hours 
of  use  the  more  efficient  pump  is  best. 

The  difference  in  total  vestances  of  two  units  operating  at 
different  efficiencies  increases  with  the  number  of  hours  of 
use.  Thus  when  N  =  5000  hours, 

Vi  =  592  +  12.5  X  5000  =  $63,092, 
and  F2  =  915  +  10.733  X  5000  =  $54,580, 

giving  a  difference  of 

63,092  -  54,580  =  $8512. 
The  difference  at  1500  hours  was  only 

19,342  -  17,015  =  $2327. 

It  is  evident  then  that  for  very  few  hours  of  use  per  year 
a  cheap  pump  should  be  used.  But  as  the  number  of  hours 
of  use  per  year  increases  not  only  should  we  use  the  more 


174  VESTANCES 

expensive  pump,  but  the  comparative  value  of  the  more  effi- 
cient unit  continually  increases.  What  we  pay  for  first  cost 
is  usually  only  a  very  insignificant  part  of  the  cost  of  service. 
The  user,  under  normal  conditions,  can  afford  to  pay  much 
more  in  first  cost  to  the  manufacturer  than  is  now  commonly 
done  for  more  efficient  units. 

115.  A  considerable  decrease  in  the  life  of  units  does  not 
greatly  increase  the  cost  of  service.     Consequently  increased 
efficiency  should  be  obtained  even  if  it  is  necessary  to  greatly 
decrease  the  life  of  the  unit.     Yet  too  often  the  reputation  of 
a  certain  make  of  apparatus  is  based  on  the  length  of  service, 
rather  than  on  the  quality  and  cost.     On  the  other  hand,  there 
is  a  tendency  to  continue  to  use  units  when  they  have  become 
inefficient  due  to  wear  or  other  causes,  so  that  the  service  is 
no  longer  rendered  economically.     Under  such  conditions,  the 
pumps  should  either  be  brought  back  to  full  efficiency,  or  else 
discarded  for  new  apparatus.     Because  under  such  conditions, 
the  excess   operating  vestance  has  become  so  great  that  the 
continued  use   of   the  worn   equipment  would   be   far   more 
expensive  than  the  cost  of  replacement. 

116.  The   vestances  of  such  items  of  a  system  as   belts, 
pipe,  and  the  like  may  easily  be  determined,  as  illustrated 
below. 

Example  34.  —  A  belt  for  a  10  h.p.  unit  costs  $20  or  $2  per 
h.p.  The  belt  efficiency  is  85%.  Assuming  power  at  one 
cent  per  h.p.h.,  interest  at  5%,  and  the  life  of  the  belt  at  5 
years,  and  neglecting  other  operating  costs,  determine  the 
vestance  per  h.p.  of  the  belt  per  normal  year  of  3000  hours. 
How  does  it  compare  with  the  vestance  of  a  flexible,  steel  link 
coupling  costing  $45? 

Solution:  With  interest  at  5%  and  the  life  at  5  years,  the 
term  factor  is  0.21649.  The  depreciation  vestance  per  h.p.  is 
then 

2  -T-  0.21649  =  $9.25. 

For  each  h.p.h.  transmitted  0.15  h.p.h.  is  lost.  The  loss  per 
h.p.  year  is  then 


PIPE  175 

3000  X  0.15  =  450  h.p.h., 
costing  450  X  o.oi  =  $4.50. 

The  operating  vestance  is  then 

4.50  -T-  0.05  =  $90, 

and  the  total  vestance  is 

9.25  +  90  =  $99.25. 
If  a  10  h.p.  flexible  coupling  costs  $45,  then  the  cost  per  h.p. 

is  $4.5°- 

For  all  practical  purposes,  the  life  of  the  coupling  is  per- 
manent.    The  depreciation  vestance  per  h.p.  is  therefore 
4.50  -r-  i  =  $4.50. 

The  efficiency  is  100%.  The  operating  vestance,  neglecting 
attendance  and  repair  as  we  did  for  the  belt,  is  then  zero  and 
the  total  vestance  is 

4.50  +  o  =  $4.50 

as  compared  with  $99.25  for  the  belt.  On  this  basis,  it  ap- 
pears therefore  that  we  could  spend  as  much  as  $94.75  per 
h.p.  for  direct  connection  and  still  render  the  service  as  cheap 
as  the  belt-driven  unit  in  cost. 

If  the  power  costs  2  cents  per  h.p.h.,  the  operating  vestance 
of  the  belt  would  be  doubled  ($180)  and  its  total  vestance  per 
h.p.  would  increase  to  $189.25.  We  could  then  afford  to  spend 
this  greater  amount  per  h.p.  for  direct  connection. 

As  a  matter  of  fact  the  difference  between  the  belt  and 
coupling  is  much  greater  than  is  indicated  above.  For, 
while  the  cost  of  attendance  and  repair,  that  we  neglected 
to  take  into  consideration  above,  is  very  small  for  the  coupling, 
it  is  very  large  for  the  belt.  Furthermore  the  belt  places  heavy 
lateral  strains  on  the  bearings  of  both  the  driver  and  driven 
unit.  The  loss  of  power  and  increased  repair  on  the  units 
that  this  occasions  must  be  charged  against  the  belt.  In  a 
well-made  coupling,  these  losses  are  eliminated,  the  power 
being  transmitted  without  such  strains. 

117.  Pipe.  —  Neglecting  attendance  and  repair  which  is 
inappreciable  in  a  well-designed  installation,  the  cost  of  opera- 


i76 


VESTANCES 


tion  of  pipe  is  the  cost  of  the  power  lost  in  transmitting  the 
fluid,  usually  water,  in  friction.  This  loss  varies  with  the 
velocity  at  which  the  fluid  is  transmitted,  increasing  rapidly 
with  increased  velocity. 

The  life  of  pipe  has  been  the  subject  of  much  argument. 
If  improperly  used,  the  life  may  be  very  short,  due  to  corro- 
sion. If,  however,  the  pipe  is  covered  with  some  noncon- 
ductor, such  as  asphaltum,  or  if  it  is  taped  and  asphalted,  this 
corrosion  is  eliminated.  On  the  whole,  it  seems  safe  to  assume 
a  life  of  50  years. 

We  give  below  the  vestances  of  pipe  based  on  5%  interest, 
cost  of  power  one  cent  per  h.p.h.  and  also  based  on  a  normal 
year  of  3000  hours.  It  must  be  borne  in  mind,  however,  that 
where  the  pipe  is  used  for  irrigation,  the  normal  year  should 

TABLE  81 

VESTANCES  OF  STANDARD  IRON  PIPE  PER  100  FT. 


SIZE, 
INCHES 
DIA. 

VELOCITY,  FT.  PER  SECOND 

0-5 

I.O 

2.0 

5.0 

10   0 

20.  o 

l 

4 
1 
1 
I 

I 

I* 

VD  .  .  . 

$2    14 

0-54 

$2.14 
ii  .46 

$2.14 

93.60 

$2.14 
660  .  oo 

$662.14 

$2.14 
810.00 

Va  

Vt  

VD 

$2.68 
$2.14 

I.  21 

$13.60 

$2.14 
15.40 

$95  74 

$2.  I^ 
106.00 
$108.14 

$3-01 
144-00 
$147-01 

$3-51 
202.50 

Va  

vt  

13-35 

$3-oi 
i-5i 

$17-54 

$3-oi 
19.30 

$812.14 

$3-01 
965.00 

VD  

Va  

$3-oi 
0.19 

Vt 

$3-20 

$3-51 
0.26 

$4-52 

$3-51 
1.87 

$22.31 

$3-51 
27.30 

$968.01 

$3-5i 
1,340.00 

VD  .                 $3  Si 

Va  0.04 

Vt  .                 $1  w 

$3-77 

$4.99 
0-39 

$5.38 

$4.99 
2.30 

$30.81 

$4.99 
31.00 

$2O6.OI 

$4.99 
232.54 

$i,343-5i 

$4-99 

i,707.75 

VD                     $4.  oo 

Va  0.05 

Vt                    $<;  04 

$5-38 

$8.23 
0.40 

$7-29 

$8.23 
3.00 

$35-99 

$8.23 
44.60 

$237-53 

$8.23 
351-00 

$1,712.74 

$8.23 
2,300.40 

VD  .    .    .    .     $8  23 

Va  0.06 

Vt              .     $8  29 

$8.63 

$11.23 

$52-83 

$359-23 

$2,308.63 

PIPE 

TABLE  81  —  Continued 


177 


SIZE, 
INCHES 
DIA. 

VELOCITY,  FT.  PER  SECOND 

o.S 

I.O 

2.O 

5-0 

IO.O 

20.  o 

2 
2j 

3 
4 
5 
6 

8 

10 
12 

V  D  •  •  •           $10  45 

$10.45 
0-51 

$10.45 
3.83 

$10.45 
56.60 

$10.45 

404.00 

$10.45 
2,767.50 

Va  0.07 

Vt  $10.52 

V  D       •  •               $17    25 

$10.96 

$l7-25 
0.64 

$14.28 
$17.25 

4-73 

$67.05 

$l7-25 
65.00 

$414.45 

$17.25 
496.12 

$513-37 

$22.70 
573-75 

$2,777-95 

$17-25 
3,213.67 

Va  0.09 

Vt  $17-34 
VD  $22.70 

Va  O.II 

$17-89 

$22.70 
o-73 

$21.98 

$22.70 
5-40 

$82.25 

$22.70 
77.20 

$3,230.92 

$22.70 
3,712.50 

Vt                            $22    8l 

$23-43 

$35-15 
0.92 

$28.10 

$35-15 
6.71 

$99  .  90 

$35-15 
92.30 

$596.45 

$35-15 
714-15 

$3,735-20 

$35-iS 
4,606  .  87 

VD  •  .  •           $35  15 

Va  O.I3 

Vt.  .           .  $•-?<;  28 

$36.07 

$48.95 
1.09 

$41.86 

$48.95 
7.80 

$127-45 

$48.95 
i  i  i  .  03 

$749-30 

$48.95 
810.00 

$4,642.02 

$48.95 
5,675-o6 

VD                   $48  95 

Va  0.16 

Vt"  .                       $4-0    II 

$50.04 
$63-25 

1.22 

$56.75 

$63  •  25 
8.91 

$159.98 
$63.25 

121  .50 

$858.95 

$63.25 
918.00 

$5,724-01 

$63-25 
6,345-QQ 
$6,408.25 

$93-70 
7,830.00 

Vn                   $6*  2$ 

Va  0.18 

Vt.  .            .   $63.43 

$64.47 

$93  •  70 
1.56 

$72.16 

$93  •  70 
ii  .50 

$184.75 

$93  -  70 
153-56 

$981  .  25 

$93  •  70 
1181.25 

Yn  .                &CH  7o 

Va  O.  2O 

Vt  $93-90 

'/  £)                        $132    OO 

$95  •  26 

$132.00 
1.86 

I5l05  .  20 

|l32.00 
13-97 

$247.26 

$132.00 
i  89  .  oo 

$1274.95 

$132.00 
1398.60 

$7,923.70 

$132.00 
10,266.75 

Va  0.24 

ft                        $1^2    24 

$133-86 

£168.00 
2.17 

$145-97 

$168.00 
I4.70 

$321.00 

$168.00 
202.00 

$1530.60 

$168.00 
1620.00 

$10,398.75 

$168.00 
10,800.00 

7D  $168.00 

7  'a  0.30 

7t  $168.30 

$170.17 

$182.70 

$370.00 

$1788.00 

$10,968.00 

be  assumed  at  only  half  of  this  amount  or  1500  hours.     Cost 
of  delivery,  laying,  etc.,  is  not  included. 

118.  Galvanized  pipe,  running  much  higher  in  first  cost 
than  the  black  pipe,  can  be  justified  only  by  a  proportionate 
increase  in  the  life  of  the  pipe,  unless  other  advantages  are 
forthcoming.  This  can  hardly  be  done.  Conducting  cover- 


178 


VESTANCES 


$1400 


1200 


1000 


Z 


800 
600 
400 
200 


6  8  10 

Diameter  in  Inches 

FIG.  34.  —  Showing  the  variation  of  total  vestance  with  the 
velocity,  and  size  of  pipe. 

ings  have  generally  been  displaced  with  nonconducting 
coverings. 

The  above  table  shows  that  at  very  low  velocity  the  operat- 
ing vestance  is  practically  negligible  in  comparison  with  the 
depreciation  vestance.  For  a  12  in.  pipe,  at  0.5  ft.  per 
second  of  velocity,  for  example,  the  depreciation  vestance  is 
$168  while  the  operating  vestance  is  only  $0.30,  or  about 
0.18%  of  the  total  vestance.  But  at  high  velocities  condi- 
tions are  reversed.  For  example,  at  20  ft.  per  second  ve- 
locity, in  the  case  of  the  12  in.  pipe,  the  operating  vestance 
is  $10,800  while  the  depreciation  vestance  is  still  $168.  The 
latter,  under  these  conditions,  is  only  1.53%  of  the  total 
vestance  and  is  thus  almost  negligible. 

119.  Somewhere  between  extremely  IOWT  and  extremely  high 
velocity  is  the  point  of  best  economy.  The  above  table  does 
not  show  this.  But  this  can  be  shown^if  the  table  is  reduced 
to  vestances  per  cubic  feet  per  second  (cu.  sec.  ft.)  of  capacity. 
This  we  have  done  in  the  table  below. 


PIPE 


179 


TABLE  82 
VESTANCES  PER  100  FT.  OF  STANDARD  PIPE  PER  Cu.  SEC.  FT.  OF  CAPACITY 


SIZE, 
INCHES 
DIA. 

VELOCITY,  FEET  PER  SECON 

0.5 

1.0 

2.O 

5-o 

10.  0 

20.  o 

i 

I 

i 

f 

i 

i| 

2 

*i 

3 

4 
5 

VD  

53150.00 

795  .  oo 

[!i  260.00 
6750.00 

$630  .  oo 

27,550.00 

$315-00 

97,000.00 

Vn 

vt  

$3945.00 

$1430.00 
717.00 

$8010.00 

$570.00 
4110.00 

$28,180.00 

$285.00 
14,140.00 

$97,315.00 
$143.00 

54,000.00 

VD 

Va  

vt 

$2147.00 

£1132.00 

566.00 

$4680.00 

$453-00 
2890.00 

$14,425.00 

$227.00 
IO,8OO.OO 

$54,143.00 

$113.00 
36,200.00 

VD  
Va  

$2265.00 
143.00 

Vt  

$2408.00 

$II7O.OO 

87.00 

$1698..  oo 

$585.00 
311.00 

$3343.00 

$234.00 
1820.00 

$11,027.00 

$117.00 
7,500.00 

$36,313.00 

$58.00 
22,350.00 

VD  .  .  .$2340.00 

Va  28.00 

Vt  $2368.00 

VD  .  .  .$1870.00 

Va  IQ.OO 

$1257.00 

$935  .00 
73.00 

$896.00 

$468.00 
159.00 

$2054.00 

$187.00 
i  160.00 

$7,617.00 

$94  •  oo 
4,360.00 

$22,408.00 

$47-00 
16,000.00 

Vt  $1889.00 
VD  .  .  .$1320.00 

Va  10.00 

$1008.00 

$660.00 
32.00 

$627.00 

$330  .  oo 
I  20.00 

$1347-00 

$132.00 
714.00 

$4,454.00 

$66.00 
2,810.00 

$16,047.00 

$33-00 
9,200.00 

Vt  $1330.00 

VD  •  .  .  $1045  -°o 

Va  7-00 

$692.00 

$522.00 
26.00 

$450.00 

$261  .00 
96.00 

$846.00 

$104.00 
566.00 

$2,876.00 

$52.00 
2,020.00 

$9,233.00 

$26.00 
6,919.00 

Vt     .     $105^  oo 

$548  .  oo 

$518.00 

19.00 

$357-00 

$259.00 
71.00 

$670.00 

$103.00 
391.00 

$2,072.00 

$51.00 
1,490.00 

$6,945.00 

$25.00 
4,670.00 

VD  .  .  .$1035.00 

Va  5-00 

Vt  $1040.00 
VD  .  .  .    $908.00 

Va  4-00 

$537-oo 

$454.00 
16.00 

$330.00 

$227.00 
54.00 

$494  .  oo 

$91.00 
309.00 

$1,541.00 

$45-oo 
1,150.00 

$4,695.00 

$23.00 
3,712.00 

Vt  $912.00 
VD  .  .  .    $803.00 

Va  3-00 

$470  .  oo 

$402.00 
15.00 

$281.00 

$201  .00 
38.00 

$400.00 
$80.00 

211.00 

$1,195-00 

$40.00 
816.00 

$3,735-00 

$20.00 
2,635.00 

Vt  $806.00 
VD  .  .  .   $734-oo 

Va  2.00 

$417-00 

$367-00 
8.00 

$239.00 

$184.00 
29.00 

$291  .OO 

$73-00 
167.00 

$856.00 

$37-oo 
607.00 

$2,655.00 

$18.00 
2,120.00 

Vt  $736.00 

$375-oo 

$213.00 

$240.00 

$644.00 

$2,138.00 

i8o 


VESTANCES 

TABLE   82  —  Continued 


SIZE, 
INCHES 
DIA. 

VELOCITY,  FEET  PER  SECOND 

0.5 

I.O 

2.0 

S-o 

10.  0 

20.  O 

6 
8 

10 
12 

VD  .  .  .    $632.00 

Va  2.00 

$318.00 

6.00 

$159.00 
22.OO 

$63.00 

122.  OO 

$32.00 
459.00 

$l6.00 
I,586.OO 
$1,602.00 

$I4.OO 
1,175.00 

Vt  $634.00 
VD  .  .  .    $562.00 

Va  1.  00 

$324.00 

$281  .00 
5.00 

$181.00 

$140.00 
17.00 

$185.00 

$56.00 
92.OO 

$491  .00 

$28.00 

355-oo 

Vt  $563.00 
VD  .  .  .    $494.00 

Va  I.  00 

$286.00 

$247.00 
4.00 

$157-00 

$148.00 
13.00 

$148.00 

$49  .  oo 
71.00 
$120.00 

$45-oo 
54.00 

$383.00 

$25.00 
262.00 

$1,189.00 

$I2.OO 

963  .  oo 

Vt  $495  -oo 
VD  .  .  .    $448.00 

Va  1.  00 

$251.00 

$224.00 
3.00 

$161  .00 

$112.00 
10.00 

$287.00 

$22.00 
216.00 

$975.00 

$11  .00 
721  .00 

Vt  $449.00 

$227.00 

$122.00 

$99  .  oo 

$238.00 

$732.00 

$2800 


2400 


7 


2  Pipe 


02000 

I 

u 

&1600 

d 


I 

I   800 


g   400 


Z 


1  Ppe 


Pipe 


12  Pipe 


8  10          12 

Velocity  in  f.p.s, 


16 


18 


20 


FIG.  35. — Total  vestances  of  standard  iron  pipe  per  100  ft.  of  length 
and  per  cu.  sec.  ft.  of  capacity. 

We  thus  see  at  a  glance  that  starting  with  low  velocity  the 
total  vestance  decreases  as  the  velocity  increases  until  it  reaches 


PIPE  181 

a  minimum,  and  thereafter  increases.  This  minimum  point  is 
the  point  of  best  economy  under  the  conditions  assumed.  If 
the  cost  of  power  is  in  excess  of  one  cent  per  h.p.h.,  lower  veloci- 
ties must  be  used,  otherwise  higher  velocities  for  best  economy. 
There  are  instances  in  the  piping  of  water  where  some  power 
is  available  but  not  sufficient  for  economical  development. 
Under  such  conditions  small  pipe  may  be  used,  wasting  this 
power  in  friction  but  reducing  thereby  the  cost  of  the  pipe. 
But  there  are  other  instances  where  the  power  is  wasted  merely 
because  of  a  lack  of  knowledge  of  its  worth.  We  have  in  mind 
one  such  instance  where  the  power  wasted  was  afterward  put 
to  use  and  found  sufficient  for  a  town  of  3000  people. 

120.  To  give  a  better  idea  of  the  variation  of  total  vest- 
ance  per  cu.  sec.  ft.,  with  velocity,  we  give  below  more  de- 
tailed data  for  the  6  inch  pipe  by  way  of  illustration. 

TABLE  83 
VELOCITY  —  FT.  PER  SECOND 

0.5        0.6      0.8      i.o     i. "5       2.0      3.0     4.0       5.0      6.0      8.0       10  15         20 

6"  VD-  $632  $527  $395  $316  $211  $159  $105  $79  $63  $53  $39  $32   $21  $16 
Va.  •   2    2   3   6   14   22   48  81  122  180  298  459  1000  1586 

Vt.  .$634  $529  $398  $322  $225  $181  $153  $160  $185  $233  $337  $491  $1021  $1602 

• 

This  shows  that  with  the.  conditions  assumed  a  velocity  of 
three  feet  per  second  is  the  point  of  best  economy  while  a 
velocity  of  four  feet  per  second  is  nearly  as  good.  At  a  velo- 
city of  20  ft.  per  second,  the  total  vestance  for  the  transmis- 
sion of  water  is  over  ten  hundred  per  cent  greater. 

The  friction  of  riveted  pipe  is  considerably  greater  than 
smooth  pipe,  such  as  standard  wrought  iron  or  wood  pipe. 
According  to  Mark's  Mechanical  Engineer's  Handbook,  that 
of  6  inch  riveted  pipe  is  1.8  ft.  per  100  ft.  of  pipe  at  3  ft.  per 
second  velocity,  while  that  of  smooth  pipe  of  the  same  size 
and  at  the  same  velocity  is  only  0.7  ft.  under  the  same  condi- 
tions; the  operating  vestance  of  the  riveted  pipe  will  be 
$122  as  compared  with  only  $47  for  the  smooth  pipe.  Since 
the  total  vestance  of  the  latter  is  only  $152,  it  would  be  neces- 
sary, for  equal  economy,  to  have  only 
152  -  122  =  $30 


182 


VESTANCES 


TABLE  84 

VESTANCES  OF  WOOD  PIPE  PER  100  FEET  PER  CUBIC  SECOND  FEET  AT  THREE  FEET 
PER  SECOND  VELOCITY 


SIZE, 
IN. 
DIA. 

HEAD  IN  FEET 

50 

IOO 

ISO 

200 

250 

300 

350 

400 

2 

3 

4 
6 
8 

10 
•  12 

16 
20 

24 

?  D    •  .  $178.00 
Va  203.00 

£182.00 
203  .  oo 

5i88.oo 
203  .  oo 

£203.00 
203  .  oo 

r2i8.oo 
203  .  oo 

1232.00 

203  .  oo 

5263.00 
203  .  oo 

£299.00 
203  .  oo 

P,....  $381.00 

VD  .  .    $87.00 
Ka....   115.00 

$385.00 

$91.00 
115.00 

$391.00 

$93-oo 
115.00 

>4o6  .  oo 

$97-00 
115.00 

£421  .00 

£101  .00 
115.00 

$435-oo 

5io8.oo 
115.00 

5466  .  oo 

£126.00 
115.00 

£502  .  oo 

$138.00 

115.00 

V  t.  .  .  .$202.00 
7D  .  .    $61.00 

Fa....       Sl.OO 

^206  .  oo 

$63.00 

81.00 

$208.00 

$64  .  oo 
81.00 

£212.00 

$78.00 
81.00 

£216.00 

$82.00 
81.00 

£223.00 

$89  .  oo 
81.00 

£241.00 

$98.00 

81.00 

$253.00 

£108.00 

81.00 

Vt  $142.00 

VD  .  .    $33  -00 
Va....     48.00 

£144.00 

$39-oo 
48.00 

5145.00 

$42.00 
48_.oo 
$90.00 

$31.00 
34-oo 

£159.00 

$46  .  oo 
48.00 

£163.00 

$50.00 
48.00 

£170.00 

$55-oo 
48.00 

$179.00 

$59-00 
48.00 

$189.00 

$61.00 
48.00 

Vt...  .   $81.00 

V  D    .  .     $22.OO 

Fa....     34-oo 

$87.00 

$28.00 
34-oo 

$94-00 

$35-00 
34-oo 

$98  .  oo 

$38.00 

.   34-oo 

£103.00 

$41.00 
34-00 

$107.00 

$44  •  oo 
34-00 

£109.00 

$51.00 
34-oo 

Ft....  $56.00 
VD  .  .    $20.00 

Fa....        26.00 

$62.00 

$26  .  oo 
26.00 

$65-00 

$28.00 
26.00 

$69  .  oo 

$32-00 
26.00 

$72.00 

$35-00 
26.00 

$75-oo 

$38.00 
26.00 

$78.00 

$41.00 
26.00 

$85.00 

$47-oo 
26.00 

Vt....  $46.00 
VD  .  .    $15-00 

Fa  ....        2O  .  OO 

$52.00 

$20.00 
20.00 

$54  •  oo 

$23.00 
20.00 

$58.00 

$25.00 
20.00 

$61.00 

$29.00 
20.00 

$64  .  oo 

$32.00 

20.00 

$67.00 

$3.7-00 
20  oo 

$73-00 

$41.00 
20.00 

Vt....  $35-00 

Fz>  .  .   $14.00 
Fa.  .  .  .     14.00 

$40.00 

$17.00 
14.00 

$43  •  oo 

$45-oo 

$49  •  oc 

$52.00 

$57-oo 

$61.00 

Vt....  $28.00 
VD 

$31.00 
$15.00 

11.00 

$26.00 

$11.00 

9.00 

Vn 

Vt  

VD  ..     $9-00 
Fa....       9-00 



Vt....  $18.00 

$20.00 

PIPE 


183 


$600 


500 


SI 


400 


5  £300 


*  a 

1C 


200 


100 


0  2  4  6  8  10  12          14  16  18          20 

Inches  Diameter  of  Pipe 

FIG.  36.  —  Total  vestances  of  wood  pipe  at  50  and  200  ft.  of  pressure. 


of  depreciation  vestance.  Assuming  then  that  the  riveted 
pipe  has  as  long  a  life  as  the  wrought  iron  pipe,  i.e.,  50  years, 
so  that  its  term  factor  will  be  0.91279,  it  would  be  necessary 
to  purchase  the  riveted  pipe  at 

30  X  0.91279  =  $27.38370  per  100  ft. 

as  compared  with  $57.80  for  the  wrought  iron  pipe,  i.e.,  the 
riveted  pipe  would  be  worth  only  47.4%  as  much  as  the  latter. 

The  use  of  extra  strong,  or  double  extra  strong,  pipe  is  war- 
ranted where,  and  only  where,  the  pressure  to  be  borne  is  too 
great  for  standard  pipe.  For  still  lower  pressures,  casing  may 
be  used  to  considerable  advantage  due  to  its  lower  first  cost. 

121.  We  give  below  the  vestances  of  wood  pipe  on  the 
same  basis  as  the  standard  pipe  above.  This  pipe  has  come 
into  very  considerable  use.  It  is  particularly  well  adapted  for 
low  heads.  It  cannot,  however,  be  allowed  to  stand  empty 
without  ruining  it.  The  friction  is  assumed  equal  to  that  of 
standard  pipe. 

Since  the  operating  vestance  of  wood  and  standard  pipe 
has  been  assumed  equal,  while  the  first  cost,  and  therefore  the 
depreciation  vestance,  of  the  former  is  much  lower,  the  total 


184  VESTANCES 

vestance  of  wood  pipe  is  far  the  lower.  However,  it  must  be 
remembered  that  there  are  some  duties  for  which  wood  pipe 
is  not  adapted.  But  in  such  service  as  the  distribution  of 
water  for  irrigation,  wood  pipe  has  proved  an  inestimable 
aid,  allowing  much  development  that  would  have  been  pro- 
hibitive with  more  expensive  pipe. 

It  would  be  a  simple  matter  in  any  concrete  case  to  com- 
pare the  total  vestances  of  pipe  with  that  of  open  ditches  or 
flumes.  However,  both  the  first  cost  and  the  cost  of  main- 
tenance of  such  conveyances  is  highly  variable,  varying  greatly 
with  the  character  of  the  ground,  the  conformation  of  the 
soil,  and  local  weather  conditions.  In  any  concrete  case, 
where  these  items  are  definitely  known,  exact  comparisons 
can  be  readily  made,  so  that  we  can  determine  which  is 
the  best. 

122.  It  must  be  borne  in  mind  that,  primarily  the  prices 
of  all  commodities  are  based  upon  the  cost  of  production 
of  these  commodities.  If  a  system  is  so  installed  that 
it  does  not  render  the  service  at  best  economy,  then  labor  is 
being  wasted.  If  the  installation  is  too  cheap  in  design  and 
construction,  then  labor  is  wasted  in  operation.  If,  on  the 
other  hand,  the  installation  is  too  expensive,  then  labor  has 
been  wasted  in  producing  a  system  more  refined  and  efficient 
than  the  amount  and  nature  of  the  service  warrants.  Waste 
of  brains,  skill  or  labor  is  a  dead  loss;  a  waste  that  no  com- 
munity, whether  state  or  nation,  can  afford.  In  the  face  of 
international  competition,  such  a  waste  leads  with  certainty  to 
national  impoverishment.  It  is  a  mark  of  incompetence. 

In  the  calculations  of  this  chapter  on  motors,  we  have  as- 
sumed a  cost  of  i  cent  per  h.p.h.  This  is,  as  we  have  pointed 
out,  far  lower  than  can  be  obtained  in  practice.  In  order  to 
indicate  what  may  actually  be  expected  we  publish  the  actual 
rates  in  Oregon  for  1919,  as  announced  by  the  public  service 
commission. 

These  are  as  follows: 


PIPE  185 


TABLE  85 
PRIMARY  RATE 


First      500  kw.h.  per  mo.  at  5  cents  per  kw.h. 

Next  500  "  "  "  "  4  "  " 

"  4,000  "  "  "  "  3  "  " 

"  10,000  "  "  "  "  2  "  "  " 

AU  excess         "  "      "     "  ij   "      "       " 


SECONDARY  RATE 

First      4,000  kw.h.  per  mo.  at  i|  cents  per  kw.h. 
Next  100,000     "       "      "     "  i.o"        "        " 
Excess  kw.h.  "      "     "0.8    "       " 


Assuming  a  load  factor  of  40%,  i.e.,  assuming  that  the 
motors  run  at  full  load  40%  of  the  time  on  the  basis  of  a  24- 
hour  day  and  are  idle  60%  of  the  time,  and  also  allowing  for 
the  unavoidable  motor  losses,  these  rates  are,  for  various  sized 
motors,  as  follows: 

TABLE  86 

Corresponding  mean  rate 
Size  of  motors  at  40%  load  factor 

Up  to  if  h.p 5  cents  per  h.p.  of  motor 

if  to  3  h.p 4*      "       "      "     "        « 

3  to  17. 5  h.p 3.3      "       "      "     "       " 

17. 5  to  52. 5  h.p 2.433      "       "      «     «       " 

52. 5  to  105  h.p 2.004     "       "      "     " 

To  determine  the  total  vestance  of  a  motor  under  such  a 
rate,  multiply  the  operating  vestance  by  the  rate  for  the  sized 
motor  under  consideration  and  add  thereto  the  depreciation 
vestance. 

For  example  in  Table  74  we  gave  for  a  one  h.p.  motor  at 
full  load: 

Depreciation  vestance $56 . 80 

Operating  vestance 767 .  oo 

Total  vestance $823 . 80 

The  operating  vestance  was  based  on  i  cent  per  h.p.h.     Ac- 


1 86  VESTANCES 

cording  to  the  above  rate  this  should  be  5  cents.     For  this 

rate,  the  operating  vestance  would  be 

767  X  5  =  $3835> 
giving  a  total  vestance  of 

3835  +  56.80  =  $3891.80 

instead  of  only  $823.80.     This  would  not  compare  well  with 
engines  of  any  type. 

We  give  in  the  table  below  the  vestances  of  induction  motors 
at  full  load  with  the  rates  as  per  the  table  above,  other  things 
as  in  the  previous  table  on  induction  motors. 


TABLE  87 

VESTANCES  PER  H.P.  OF  INDUCTION  MOTORS  AT  FULL  LOAD  AND  40%  LOAD 

FACTOR 

AS  PER  THE  1919  .OREGON  RATE 


SIZE  OF  MOTOP, 
H.P. 

VESTANCES 

Depreciation 

Operating 

Total 

I 

$56.80 

$3835-00 

$3891.80 

2 

35-55 

3276.00 

33II-55 

3 

26.95 

2353-00 

2379-95 

5 

19.80 

2310.00 

2329.80 

7l 

25.60 

2273.70 

2298.30 

10 

22.75 

2257.20 

2279.95 

15 

18.20 

2234.10 

2252.30 

20 

16.70 

1640.  10 

1656.80 

25 

14.50 

1632.80 

1647.30 

30 

14-45 

1627.90 

1642.35 

40 

12.43 

1623.00 

I635-43 

SO 

n-37 

1618.  20 

1629.57 

75 

9.76 

1326.00    . 

I335-76 

CHAPTER   VI 
UNIT  COST  DETERMINATION 

Unit  Cost.  Time  Element.  Change  Points.  Unit  Cost  at  Constant 
Load.  Nearly  Constant  Load.  Two  Part  Load.  Three  Part  Load. 
N-part  Load.  Instantaneous  Unit  Cost  of  Actual  Load  Curve. 
Service  Modulus  and  Replot. 

123.  We   have   seen   in   the   foregoing   chapters   that   cost 
governs  price,  so  that  our  real  problem  is  primarily  that  of 
cost  determination.     But  in  what  way  shall  we  apportion  the 
costs,  when  all  elements  that  enter  into  the  problem  are  vari- 
able?    The  answer  to  this  is:    That  an  equipment  must  earn  its 
costs,  and  profit,  while  in  operation  — for  certainly  it  cannot  do 
so  while  idle.     This  is  the  fundamental  basis  of  cost  analysis. 

124.  Let  us  then,  to  begin  with,  take  the  simplest  condi- 
tions of   operation  possible,   that  in  which  the   undertaking 
runs  at  full  load  for  a  variable  number  of  hours  per  year. 

Let 

C  =  the  cost  per  h.p.  of  the  system, 
p  =  the  per  cent  of  fixed  charges, 
a  =  the  operating  cost  per  h.p.h., 
L  =  the  load  carried  and 
N  =  the  number  of  hours  of  operation 

then,  evidently 

pC  =  the  fixed  charges  per  h.p.  year 
and  LpC  =  the  total  fixed  charges  per  year. 

The  greatest  value  that  (N)  can  have  is  8760,  the  number 
of  hours  in  a  year.  Again  since  the  operating  cost  is  assumed 
to  be  constant  at  (a)  dollars  per  h.p.h.  (or  other  unit  of  service), 
then  La  =  the  total  cost  of  operation  per  hour, 

and  LaN  =  the  total  cost  of  operation  per  year. 

187 


i88  UNIT  COST   DETERMINATION 

Evidently  then  the  total  production  cost  (Ki)  per  year  is 
the  sum  of  the  total  annual  cost  of  operation  and  the  fixed 
charges  or 

Ki  =  LpC  +  LaN  =  L(pC  +  aN)  .    .    .    (69) 

If  now  we  divide  this  total  annual  cost  by  the  load  (L) 
and  the  hours  of  operation  (N),  we  will  get  the  cost  per  h.p.h. 

(#2)or  Ki        L(pC  +  aN) 

KZ~TN  =      ~~LN~ 

•   S~1 

or  K2  =  ~  +  a  ............    (70) 

As  (N)  becomes  less,  the  unit  cost  (K2)  becomes  greater, 
until  when  N  =  o,  the  unit  cost  is  infinite,  the  operating  cost 
per  h.p.h.  being  assumed  constant.  On  the  other  hand  as  (  N) 
increases,  the  unit  cost  becomes  less,  tending  to  decrease  to 
the  value  of  the  operating  cost,  when  (N)  =  »  .  But  the 
largest  value  that  (  N)  can  have  is  8760,  the  number  of  hours 
in  the  year,  and  therefore  the  minimum  value  that  the  unit 
cost  can  reach  is 


a 

8760 

when  the  plant  is  operated  at  full  load  throughout  the  entire 
year. 

125.  If  we  have  two  plants,  one  of  which  has  a  first  cost  of 
Ci  dollars,  a  depreciation  of  pi  per  cent,  and  an  operating  cost 
of  #i  dollars,  while  the  second  has  corresponding  values,  C2, 
pz,  and  #2  and  each  carry  the  same  load  L,  then  the  total 
annual  cost  of  the  first  plant  is 


and  of  the  second  is 


If  we  plot  the  total  annual  production  cost  vertically  as  in 
Fig.  37  and  the  hours  horizontally,  then  these  equations  will 
give  straight  lines  AB  and  A'B'.  It  may  be  and  often  is  true 
that  two  such  lines  as  AB  and  A'B'  intersect  between  N  =  o 


THE   CHANGE   POINT 


Hours  of  Operation  per  Year 


8760 


FIG.  37.  —  Variation  of  production  cost  with  use  and  the  change  point. 

and    N  =  8760,  as  at  (C),  Fig.  37.     At    this  change    point 
Ki  and  K"  are  equal  and  therefore 


or  piCi  +  01  N  =  p2C2  +  a2N, 

and  solving  for  N  gives  the  change  point 


(71) 


For  values  of  (N)  less  than  N0,  Ki  is  less  than  Ki"  For 
values  of  (N)  greater  than  N0,  Ki"  is  less  than  Ki.  Evi- 
dently then  when  the  hours  of  operation  are  less  than  A^0,  the 
first  plant  is  the  most  economical,  for  a  greater  number  of 
hours  of  operation,  the  second  plant  is  best.  The  application 
of  this  will  be  illustrated  in  the  problems  below. 

Example  35.  —  A  pumping  plant,  having  a  capacity  of  ten 
thousand  gallons  per  minute  costs  $8000  and  has  14%  fixed 
charges.  The  cost  of  operation  of  the  plant  is  20  cents  per 
hour.  What  is  the  cost  per  hour  per  1000  gallons  of  capacity? 

Solution:  Under  the  above  conditions  the  fixed  charges  are 
8000  X  14%  =  $1120, 


UNIT  COST  DETERMINATION 


or  $ii2  per  1000  gallons  of  capacity.  The  total  cost  of  opera- 
tion is  20  cents  per  hour  or  2  cents  per  1000  gallons  of  capacity. 
For  (AO  hours  this  cost  would  be  o.o2Ar  dollars. 

The  total  production  cost  per  thousand  gallons  of  capacity 
per  year  is  then 

KI  =  112  +  O.O2N (a) 

The  cost  per  hour  is  obtained  by  dividing  (K\)  by  the  num- 
ber of  hours  (N),  getting 

K     =  — 

2  =     N 


0.02 


We  can  get  values  for  K\  and  K^  from  equations  (a)  and  (b) 
for  various  values  of  (N)  as  follows; 


TABLE 


N 

Kt 

K, 

o 

$112.00 

infinite 

IOO 

114.00 

$1.14 

1000 

132.00 

0.132 

3000 

172.00 

°-°573 

5000 

2  I  2  .  OO 

0.0424 

7000 

252.OO 

o  .  0360 

8760 

287.  20 

0.03278 

Plotting  the  values  of  KI  vertically  against  the  time  ( A^) 
horizontally  as  in  Fig.  38  gives  a  hyperbola.  Note  the  enor- 
mous variation  of  the  unit  cost  (Kz)  with  time.  The  number 
of  hours  of  operation  per  year  is  the  most  important  factor  in 
the  cost  of  a  service. 

Example  36.  —  A  power  plant  costs  $100  per  h.p.  and  is  of 
2000  h.p.  capacity,  running  at  full  load.  If  the  fixed  charges 
are  15%  and  the  cost  of  operation  0.5  cent  per  h.p.h.  (a)  what 
is  the  total  annual  cost;  and  (b)  what  is  the  cost  per  h.p.h.? 

Solution:  The  total  cost  of  the  plant  is 

2000  X  ioo  =  $200,000, 
and  the  total  fixed  charges  are 

200,000  X  .15  =  $30,000. 


CONSTANT   LOAD 


191 


The  operating  cost  per  year  is 

2000  X  0.005  X  N  =  10  N    .    .    . 
The  total  annual  production  cost  per  h.p.  is  then 


$1.40 


-         =    r  +  0.005 

2000  N          N 


(fr) 


1000       2000      3000       4000       5000       6000       7000       8000    8760 
Hours  per  Year 

FIG.  38.  —  Variation  of  total  and  unit  production  cost  with  use, 
according  to  example  35. 

The  values  of  KI  and  K%  for  various  values  of  (N)  are 
given  below,  in  order  again  to  illustrate  the  enormous  varia- 
tion in  cost  with  the  time  of  operation. 


TABLE   89 


.V 

Ki 

Kt 

o 

$30,000  .  oo 

infinite 

IOO 

31,000.00 

o-i55 

IOOO 

3000 
5000 

40,000.00 
60,000  .  oo 
80,000  .  oo 

O.O2 
O.OI 

0.008 

7000 

8760 

100,000.00 
117,600.00 

0.007143 
O.OO67I2 

192  UNIT  COST  DETERMINATION 

Note  that  for  1000  hours  of  operation,  the  cost  per  h.p.h.  is 
30  times  as  great  as  for  continuous  ( N  =  8760  hours)  opera- 
tion during  the  year. 

Example  37. — A  pumping  plant  of  1000  g.p.m.  capacity 
costs  $2400.  The  fixed  charges  are  12  %  and  the  cost  of  opera- 
tion is  0.3  cent  per  thousand  gallons  delivered.  Each  acre 
irrigated  requires  43,200  gallons  of  water  per  month,  so  that 
with  12  hours  of  operation  per  day,  we  can  irrigate  50  acres. 
We  wish,  however,  to  irrigate  100  acres,  and  in  as  much  as 
night  irrigation  is  not  feasible,  the  question  is,  shall  we  put  in 
another  pumping  unit,  or  shall  we  build  a  reservoir,  pumping 
into  it  during  the  night?  If  we  put  in  the  reservoir,  the 
pumping  plant  will  operate  24  hours  per  day  and  therefore 
the  power  company  will  reduce  our  rates  so  that  the  cost  of 
the  water  will  be  reduced  to  0.2  cent  per  1000  gallons  of 
capacity.  The  reservoir  costs  25  cents  per  yard  to  excavate 
and  the  land  costs  $200  per  acre.  The  reservoir  permits  of 
only  10  ft.  depth  so  as  to  keep  the  bottom  above  the  land 
level.  The  fixed  charges  on  the  reservoir  are  8%  and  the 
length  of  the  irrigation  season  is  5  months.  Shall  we  install 
another  pumping  plant,  or  put  in  the  reservoir? 

Solution:  The  question  can  only  be  answered  by  determin- 
ing the  costs  in  each  case  and  comparing  them,  the  one  ren- 
dering the  service  cheapest  being  the  one  we  shall  decide  upon. 

Where  we  are  operating  12  hours  per  day  our  costs  are  as 
follows : 

Fixed  charges  per  year  are 

2400  X  0.12  =  $288, 
and  the  operating  costs  are 

$0.003  X  60  =  $0.1 8  per  hour, 

and  since  5  months  =  5X30X12  =  1800  hours  the  annual 
cost  of  operation  is 

.18  X  1800  =  $324. 

The  total  annual  cost  is  then 

K\  =  288  +  324  =  $612  per  year. 


CONSTANT   LOAD  193 

If  we  install  two  units,  this  cost  will  be  doubled,  or 
Ki   =  2  X  612  =  $1224. 

To  hold  the  water  pumped  during  12  hours,  the  capacity  of  the 
reservoir  will  have  to  be 

1000  X  60  X  12  =  720,000  gal. 
or  720,000  -r-  7.5  =  96,000  cu.  ft., 

or  96,000  -T-  27  =  3556  yards, 

costing  3556  X  0.25  =  $889. 

If  the  reservoir  is  10  ft.  deep,  it  will  cover 
96,000  -j-  10  =  9600  sq.  ft., 

which,  allowing  for  some  wastage  of  ground  will  use  up  about 
^  acre,  costing 

200  X  i  =  $100, 

so  that  the  total  reservoir  costs  are 

889  +  100  =  $989, 
on  which  the  fixed  charges  are 

989  X  0.08  =  $79.12  per  year. 

With  one  pumping  unit  in  combination  with  the  reservoir,  the 
total  fixed  charges  per  year  are 

288  +  79.12  =  $367.12 
and  the  operating  costs  will  be,  at  the  reduced  power  rates, 

0.002  X  60  =  $0.12  per  hour, 
or  0.12  X  3600  =  $432, 

five  months  of  operation  amounting  to  3600  hours  now,  based 
on  24  hours  per  day  of  operation. 
The  total  annual  costs  are  then 

367.12  +  432  =  $799.12. 

To  summarize  then, 

(a)  With  two  pumping  units  running  12  hours  per  day  for 
5  months,  each  of  1000  g.p.m.  capacity,  the  total  annual  cost 
is  $1224. 

(6)  With  one  pumping  unit  of  1000  g.p.m.  capacity  running 
24  hours  per  day  for  5  months,  and  thus  pumping  an  equal 


i94  UNIT  COST  DETERMINATION 

amount  as  the  two  units  under  (a)  above,  in  combination  with 
the  reservoir,  the  total  annual  costs  are  only  $799.12,  showing 
a  saving  of 

1224  —  799.12  =  $424.88  per  year 
or  424.88  -i-  799.12  =  53%. 

126.  Example  38.  —  A  centrifugal  pumping  unit  of  1000 
g.p.m.  capacity  costs  $1100  and  is  guaranteed  to  operate  at 
60%  efficiency.  A  triplex  plunger  pumping  unit  of  the  same 
size  costs  $2400  and  is  guaranteed  at  75%  efficiency.  If  in 
either  case  the  fixed  charges  are  15%  and  the  power  charges 
are  i  cent  per  h.p.h.,  other  operating  costs  being  i  cent  per 
hour,  which  equipment  should  we  use,  the  head  in  either  case 
being  80  ft.? 

Solution:   (a)  In  the  case  of  the  centrifugal  pump: 
The  fixed  charges  are     noo  X  0.15  =  $165  per  year, 

.  ,  .    1000  X   80  , 

the  power  consumed  is  4ooo  x  p  6p  =  33-3  h-P, 

The  cost  per  hour  for  power  is  then 

33.3  X  o.oi  =  $0.333, 
and  the  total  operating  cost  per  hour  is 

°-333+  o-oi  =  $0.343, 

or  °-343  N  Per  year  of  N  hours. 

The  total  annual  cost  is  therefore 

K!  =  165  +  0.343^. (a) 

(b)  In  the  case  of  the  triplex  pump: 
the  fixed  charges  are        2400  X  .15  =  $360, 

,  .      1000  X  80          *     « 

the  power  consumed  is =  26.7  h.p., 

4000  X  0.75 

so  that  the  cost  per  hour  for  power  is 

26.7  X  o.oi  =  $0.267, 

or  0.277  N  per  year  of  N  hours. 

The  total  annual  cost  is  therefore 

KS  =  360  +  0.277^ :,  (b) 


.       CONSTANT  LOAD  195 

For       N  =  1000  hours  per  year, 

Ki  =  165  +  343  =  $508  for  the  centrifugal  pump, 
and    KI   =  360  +  277  =  $637  for  the  triplex  pump,  showing 
the  centrifugal  pump  to  advantage. 

For       N  =  5000  hours  per  year, 

then    Ki  =  165  +  1715  =  $1880  for  the  centrifugal  pump, 
and    Ki    =  360  +  1385  =  $1745  for  the  triplex  pump,  show- 
ing the  advantage  to  rest  with  this  pump. 

We  can  determine  for  what  number  of  hours  per  year  the 
two  costs  will  be  equal,  by  equating  (a)  and  (6),  thus 

165  -f  0.343^1  =  360  +  0.277^1 
so  that  0.066  Ni  =  195 

or  Ni  =  2955  hrs.  per  year. 

That  is,  if  we  operate  less  than  2955  hours  per  year,  then  under 
the  conditions  assumed  above,  the  centrifugal  pump  is  best; 
if  we  operate  more  than  2955  hours  per  year,  then  the  triplex 
pump  is  best. 

Example  39.  —  We  wish  to  generate  power  under  full  load 
conditions  and  find  that  a  hydro-electric  plant  will  cost  $100 
per  h.p.  and  the  cost  of  operation  will  be  o.i  cent  per  h.p.h. 
A  Diesel  engine  plant  will  cost  $60  per  h.p.  and  cost  0.2  cent 
per  h.p.h.  for  operation,  while  a  steam  plant  will  cost  $45  per 
h.p.  and  0.3  cent  per  h.p.h.  to  operate.  If  the  fixed  charges 
in  all  cases  above  are  14%  for  what  hours  of  operation  would 
each  of  the  above  plants  be  best? 

Solution:  The  total  annual  fixed  charges  per  h.p.  for  the 
hydro-electric  plant  are 

100  X  0.14  =  $14. 
For  the  Diesel  plant,  they  are 

60  X  0.14  =  $8.40, 
and  for  the  steam  plant,  they  are 

45  X  0.14  =  $6.30. 

The  annual  cost  of  operation  for  the  hydro-electric  plant  per 
h.p.  is  $0.001  N.  For  the  Diesel  plant,  it  is  $0.002  N,  and  for 


196  UNIT   COST   DETERMINATION 

the  steam  plant,  it  is  $0.003  N,  where  in  each  case  N  is  the 
number  of  hours  of  operation  per  year.  The  total  annual 
cost  per  h.p.  for  the  hydro-electric  plant  is  then 

KI  =  14.00  +  o.ooiN. 
For  the  Diesel  plant,  it  is 

KI    =  8.40  +  o.oo2N, 
and  for  the  steam  plant,  it  is 

KI"  =  6.30  +  0.003^. 
Equating  KI  and  KI',  we  get 

6.30  +  0.003^1    =   8.40  +  0.002^1, 

so  that  o.oo i Ni  =  2.10, 

and  N!  =  2100  hours  per  year. 

So  also  equating  KI  and  KI,  we  get 

8.40  +  0.002^2  =  14.00  +  o.ooi7V2 
so  that  o.ooiA^  =  5.60 

or  N%  =  5600  hours  per  year. 

The  conclusions  then  are: 

(a)  If  the  plant  is  to  operate  less  than  2100  hours  per  year, 
use  the  steam  plant. 

(b)  If  the  plant  is  to  operate  more  than  2100  hours  per  year, 
but  less  than  5600  hours  per  year,  use  the  Diesel  plant. 

(c)  If  the  plant  is  to  operate  more  than  5600  hours  per  year, 
use  the  hydro-electric  plant. 

127.  If  the  plant  does  not  operate  at  full  load  but  at  a  con- 
stant fractional  load,  then  calling  M  the  capacity  of  the  plant, 
and  L  the  load,  as  before,  we  get  that  the  annual  fixed  charges 
are  MCp,  the  annual  operating  cost  is  LaN,  the  total  annual 

cost  is 

KI  =  MCp  +  LaN, 

and  the  cost  per  h.p.h.  is 

KI       MCp  ,  f    , 

•   -.  -'-         *-Zff--ffL+fl    (72) 

and  the  application  is  as  above. 


PROBLEMS  197 

Example  40.  —  A  steam  generating  plant  running  at  80  % 
capacity  costs  $50  per  h.p.  If  the  fixed  charges  are  12%  and 
the  operating  costs  are  0.3  cent  per  h.p.h.,  what  will  be  the 
annual  cost  per  h.p.  and  the  cost  per  h.p.h.? 

Solution:  Here  M  =  i,  L  =  0.8,  p  =  0.12,  a  =  $0.003  an<3 
C  =  $5°, 

whence                      KI  =  i  X  50  X  0.12  +  0.8  X  0.003 A" 
or  KI  =  6  +  0.0024N (a) 

11                         TT        i  X  50  X  12    , 
and  also  Kz  = ^ h  0.003 

or  Kz  =  ^  +  0.003 (b) 

While  in  the  above  we  have  shown  the  application  of  the 
cost  analysis,  under  special  conditions,  only  to  pumping  and 
power  generation,  it  is  equally  applicable  to  railroad  or  wagon 
road  analysis  as  to  dry  goods  or  grocery  business. 

PROBLEMS 

1.  A  pumping  plant  of  2000  g.p.m.  capacity  costs  $2100.     The  fixed 
charges  are  13%,  and  the  cost  of  operation  is  35  cents  per  hour.    The  plant 
operates  for  2000  hours  per  year  at  full  load. 

(a)  What  is  the  total  annual  cost  of  operation  per  year? 

(b)  What  is  the  total  cost  of  operation  per  hour  per  thousand  g.p.m.  of 
capacity? 

2.  If  in  problem  (i),  the  plant  operates  for  (N)  hours  per  year: 

(a)  What  is  the  total  cost  of  operation  per  year? 

(b)  What  is  the  total  cost  of  operation  per  hour  per  thousand  g.p.m.  of 
capacity? 

(c)  Plot  the  curves. 

3.  An  electric  pumping  unit  of  10  h.p.  size  costs  $240,  while  a  gas  engine 
driven  unit  costs  $600  complete.    The  fixed  charges  in  either  case  are  12%. 
The  cost  of  operation  of  the  electric  pumping  unit  is  2  cents  per  h.p.h., 
while  that  of  the  gas  engine  driven  unit  is  1.2  cents  per  h.p.h.    Determine 
for  what  number  of  hours  of  operation  at  full  load  each  is  best.    Plot  the 
curves. 

4.  A  10  h.p.  gas  engine  electric  lighting  unit  costs  $90  per  h.p.  and  runs 
at  full  load  for  3000  hours  per  year.    The  load  increases  during  the  period 
of  operation  to  15  h.p.    We  can  take  care  of  this  additional  load,  either  by 
installing  another  5  h.p.  unit  as  above,  or  by  putting  in  a  storage  battery 


198  UNIT  COST  DETERMINATION 

and  running  the  10  h.p.  unit  4500  hours  per  year.  The  battery  costs  $75 
per  kw.  of  capacity  and  has  an  efficiency  of  65%.  The  fixed  charges  on  the 
above  units  are  15%.  The  operating  cost  of  the  engine  unit  is  1.3  cents 
per  h.p.h.  and  of  the  battery  0.2  cent  per  kw.h.  Should  we  install  the 
extra  engine  unit  or  the  storage  battery  for  best  economy? 

5.  In  problem  (4),  if  the  load  increased  to  20  h.p.,  which  should  we 
install? 

6.  In  problem  (4),  if  the  load  increased  only  to  12  h.p.,  which  should  we 
install  for  best  economy? 

7.  Given  the  following  data:    A  horizontal  4-valve  engine  generating 
unit  costs  $22,700.     The  engine  runs  with  steam  at  150$  pressure,  100° 
superheat  and  with  a  26  in.  vacuum,  and  uses  16.5$  steam  per  kw.h.    A 
steam  turbine  generating"  unit  of  the  same  size  costs  $12,250  and  operates 
under  the  same  conditions  as  the  steam  engine  unit  except  with  a  28  in. 
vacuum.    It  consumes  17.7$  steam  per  kw.h.    The  plants  are  run  at  full 
load.     If  the  fixed  charges  on  either  plant  are  14%  and  the  steam  costs 
18  cents  per  looojf,  and  other  operating  costs  are  0.06  cent  per  kw.h.,  for 
what  length  of  service  per  year  is  each  best  adapted? 

8.  A  hydro-electric  power  plant  costs  $120  per  kw.h.  and  operates  at 
60%  of  full  load,  for  (N)  hours  per  year.     Determine: 

(a)  The  total  annual  cost  per  kw.  of  the  service; 

(b)  The  cost  per  kw.h.  of  the  service; 

(c)  Plot  the  curves  for  (a)  and  (b). 

9.  We  find  that:   a  hydro-electric  plant  will  cost  $100  per  kw.  to  install 
and  0.2  cent  per  kw.h.  to  operate;   a  Diesel  plant  will  cost  $70  per  h.p. 
to  install  and  0.3  cent  to  operate;   while  a  steam  plant  will  cost  $50  per 
h.p.  to  install  and  0.4  cent  to  operate.    The  fixed  charges  on  the  hydro- 
electric plant  are  14%,  on  the  Diesel  plant  15%  and  16%  on  the  steam 
plant.    The  plants  are  to  operate  at  full  load.    For  what  number  of  hours 
of  operation  is  each  of  the  above  plants  best? 

10.  A   pumping  plant  of    1000  g.p.m.  capacity  delivers  water  against 
120  ft.  head  with  a  guaranteed  efficiency  of  70%  at  full  load  and  40%  effi- 
ciency at  half  load.    The  plant  costs  $850  to  install.    The  power  costs  one 
cent  per  h.p.h.,  while  fixed  charges  are  12%.    The  plant  operates  at  full 
load  for  2000  hours  per  year  and  at  half  load  for  1000  hours  per  year.    Deter- 
mine the  total  annual  cost  of  operation. 

11.  If  in  problem  (10),  the  plant  operated  for  (Ni)  hours  per  year  at  full 
load  and  for  (7V2)  hours  per  year  at  half  load,  what  would  be  the  total  annual 
cost  of  operation? 

12.  In  problem  (n),  plot  the  curve  for  the  variation  of  the  total  annual 
cost  of  production  with  (Ni)  if  Ni  +  Nz  =  3000  hours. 

13.  We  are  irrigating  100  acres  of  land  for  5  months  in  the  year  with  a 
25  h.p.  direct  connected  electric  centrifugal  pumping  unit  costing  $2200. 
The  pump  efficiency  is  60%,  motor  efficiency  85%,  while  the  head  pumped 


PROBLEMS  199 

against  is  40  ft.  total.  The  electric  power  costs  1.5  cents  per  kw.h.  The 
plant  runs  at  full  load  for  8  hours  each  day.  We  wish  to  irrigate  300  acres 
of  land  and  can  do  so  with  the  unit  already  installed  by  running  24  hours 
per  day,  in  which  case  the  power  will  cost  us  only  one  cent  per  kw.h.  But 
in  that  case,  we  must  install  a  reservoir  to  avoid  night  irrigation,  costing 
35  cents  per  cubic  yard  to  excavate  and  having  a  maximum  depth  of  12  ft. 
The  land  costs  $100  per  acre.  Fixed  charges  on  the  pumping  unit  are  12%, 
and  other  operating  costs  are  3  cents  per  hour.  The  fixed  charges  on  the 
reservoir  are  10%,  with  an  annual  cost  of  $78  for  maintenance  and  repair. 
With  the  reservoir  in  use  the  head  on  the  pump  will  be  increased  by  4  ft., 
with  a  corresponding  increase  in  power  consumed. 

Instead  of  using  the  above  layout,  we  can  install  two  more  25  h.p.  pump- 
ing units,  the  three  units  operating  for  8  hours  per  day,  or  we  can  install 
one  more  such  25  h.p.  pumping  unit  together  with  sufficient  reservoir  to 
avoid  more  than  8  hours  of  operation.  In  the  case  of  the  three  25  h.p. 
units,  our  power  cost  will  be  1.3  cents  per  kw.h.  and  1.2  cents  for  the  two 
25  h.p.  units.  In  the  latter  case  the  head  will  be  increased  by  2  ft.,  with 
a  corresponding  increase  in  power  consumed. 

Determine  which  of  the  three  possible  types  of  installation  are  best,  and 
by  what  per  cent. 

14.  A  10  h.p.  direct  connected  centrifugal  pump  operates  at  60%  effi- 
ciency at  full  load,  for  2000  hours  per  year.    We  wish  to  operate  for  another 
1000  hours  per  year  with  a  load  of  5  h.p.    If  we  use  the  above  pump,  its 
efficiency  will  drop  to  45%,  while  if  we  install  a  5  h.p.  unit  it  will  give 
52%  efficiency.    The  power  costs  2  cents  per  h.p.h.  while  fixed  charges  are 
12%.    Should  we  install  the  additional  unit  or  use  the  10  h.p.  unit  at  half 
load?    By  what  per  cent  would  the  one  exceed  the  other?    The  current  is 
a.c.  60  cy.  3  phase  220  volts. 

15.  We  wish  to  install  a  10  h.p.  electrically  driven  centrifugal  pumping 
plant.    Which  would  be  best,  (a)  a  belt  driven  unit  or  (b)  a  direct  connected 
unit?     Take  into  consideration  cost  of  units,  building  and  foundations. 
How  much  would  we  lose  by  installing  the  poorer  one? 

Assume  a  power  cost  of  2  cents  per  kw.h.,  fixed  charges  10%  and  1500 
hours  per  year  of  operation  at  full  load  pumping  against  a  50  ft.  head. 

16.  Determine  the  same  as  for  problem  (15)  but  for  a  100  h.p.  plant 
with  a  power  cost  of  i  cent  per  kw.h.,  other  things  being  equal. 

17.  Determine  the  same  as  for  problem  (15)  but  for  a  one  h.p.  plant 
with  a  power  cost  of  3  cents  per  kw.h.,  other  things  being  equal. 

18.  Determine  the  same  as  for  problem  (15)  but  for  a  50  h.p.  plant  with  a 
power  cost  of  1.5  cents  per  kw.h.,  other  things  being  equal. 

19.  A  250  h.p.  pumping  plant  is  to  operate  against  a  head  of  100  ft. 
Into  how  many  units  should  we  divide  this  plant  for  best  economy,  taking 
into  account  cost  of  building  and  foundations?    Power  cost  one  cent  per 
kw.h.,  hours  of  operation  2000  per  year  and  fixed  charges  10%. 


200 


UNIT  COST  DETERMINATION 


20.  Determine  the  same  as  in  problem  (19)  for  a  150  h.p.  plant  operating 
against  50  ft.  head,  other  things  being  the  same. 

21.  Determine  the  same  as  in  problem  (20)  but  for  a  30  h.p.  plant  operat- 
ing against  25  ft.  head. 

128.    Thus  far  we  have  considered  an  equipment  operating  at 
one  uniform  load  during  the  entire  year.     If  the  load  varies  with 

the  time,  as  shown  by  the 
curve,  OO,  Fig.  39,  but  not 
sufficiently  so  to  necessitate 
the  discontinuance  of  opera- 
tion of  part  of  the  plant, 
then  for  any  given  period 
we  determine  the  mean  load 
(Lm),  and  base  our  cost 
analysis  thereon,  instead  of 
0  ^  on  the  actual,  variable  load 

(M).  In  an  electric  gener- 
ating plant,  the  kw.h.  produced,  divided  by  the  time  in  hours, 
gives  the  mean  load.  In  general,  the  total  production  divided 
by  the  time  in  hours 
gives  the  mean  load  or 
average  production  per 
hour,  i.e.,  the  mean  rate 
of  production. 

But  now,  suppose  we 
have  a  twin  load  as  illus- 
trated in  Fig.  40,  i.e.,  a 
load  of  (Lz  +  Li)  for  (A2) 


-Nr 


hours  per  year  and  follow-   o 

ing  that,  a  load  of  (Li) 

for  (Ni  —  Nz)  hours  per     te- Nt- 

year.      It  is  Our  problem  FIG.  40.  — Diagram  for  a  two-part  load. 

then   to    determine   the 

cost  of  production  for  a  given  plant,  already  in  operation,  or  else 
the  more  difficult  and  important  problem  to  so  design  the  plant 
that  the  production  will  be  at  a  minimum  cost.  In  this  case  the 


TWO-PART  LOAD  201 

difference  between  L%  and  LI  is  assumed  to  be  so  great  that  it 
would  be  uneconomical  to  operate  the  entire  plant  for  the 
lesser  load,  the  entire  plant  operating  for  the  load  (L\  +  Lz) 

and   (  7  —  ^—j-  )  per  cent  operating  for  the  load  LI. 
\LI  -f-  LZ/ 

The  difficulty  in  the  solution  of  this  problem  comes  in  the 
equitable  distribution  of  fixed  charges,  primarily.  But  this 
difficulty  is  removed  if  you  bear  in  mind  that  an  equipment 
or  unit  thereof  must  earn  its  costs  and  profits  while  in  opera- 
tion, for  certainly  it  cannot  do  so  when  idle.  Suppose  then 
that  this  plant  consists  of  two  units,  one  of  (Li)  size  and  the 
other  of  (L2)  size.  Evidently  then  the  unit  (Li)  will  operate 
for  (Ni)  hours  per  year,  while  the  unit  (Lz)  operates  for  only 
(Nz)  hours  per  year.  During  the  balance  of  the  year,  above 
(Ni)  hours,  i.e.,  for  A^0  hours  (where  Ni  +  N0  =  8760  hours) 
the  load  is  assumed  to  be  zero. 
For  the  unit  (Li)  we  have 

LiCp  =  total  fixed  charges  per  year 
and      LidNi  =  total  operating  costs  per  year, 
where        LI  =  the  load, 

C  =  cost  of  plant  per  unit  of  load, 
p  =  per  cent  of  fixed  charges, 
a  =  operating  cost  per  hour  per  unit  of  load 
and  Ni  =  number  of  hours  of  operation. 

Whence  the  total  annual  cost  (  Kn)  is 

Kn  =  LiCp  4-  L&Ni 
and  the  cost  per  unit  of  load  (h.p.h.,  kw.h.  etc.)  per  hour  (^21) 

is  Kn  =  Q  +  a.  ' 

So,  also,  if  only  the  unit  (Lz)  were  in  operation  for  the 
hours,  then  the  total  annual  cost  KU  being 


the  cost  per  unit  of  load  per  hour  is 


202  UNIT  COST  DETERMINATION 

The  total  cost  for  operating  the  unit  (L2)  for  (N2)  hours  is 
Ki2    =  L^Cp  +  I^aN2 

and  for  operating  the  unit  (Li)  for  N2  hours  the  total  cost  is 
obtained  by  multiplying  the  cost  per  hour  (Eq.  68)  by  the  load 
(Li)  and  the  hours  N2)  i.e.,  by  LiA^,  getting 

,       L,N2Cp 

Ai2v  =  — --—     -f-  Lial\2. 

The  total  cost  of  both  units  operating  for  A"2  hours  is  then 
the  sum  of  Kn  and  Ki2ff  or 

Kn  =  LtCp  +  -   ^ 
i.e., 


and  the  cost  per  unit  of  load  per  hour  is  obtained  by  dividing 
by  the  total  load  (Li  +  L2)  and  the  hours  AT2  or 

/L2N1  +  L1N2\      Cp 

~  V      N,N2 


,    . 
a    '    '(73) 

for  the  period  (N2),  while  for  the  period  (N\  —  N2)  it  was,  by 
equation  (68) 

„        Cp   . 

^  =  ^+a 

and  nothing  for  the  period  (N0)  ,  since  it  is  useless  for  this  period 
according  to  the  assumptions  made  in  the  problem. 

129.  Example  40.  —  A  steam  power  plant  costs  $50  per  h.p. 
and  is  of  2000  h.p.  size.  It  carries  a  load  of  2000  h.p.  for  3000 
hours  of  the  year  and  500  h.p.  for  an  additional  2000  hours. 
If  the  cost  of  operation  is  0.3  cent  per  h.p.h.  and  fixed  charges 
are  12%,  determine  the  cost  per  h.p.h.  for  each  period,  the 
load  being  considered  integral. 

Solution:  We  can  determine  the  costs  by  substitution  in 
the  equations,  or  work  them  out  in  detail  as  follows: 


TWO-PART  LOAD  203 

The  500  h.p.  unit  will  operate  for 

3000  H-  2000  =  5000  hours. 

The  annual  operating  cost  of  this  unit  is 
500  h.p.  X  $0.003  X  5000  hrs.  =  $7500, 

and  the  fixed  charges  are 

500  h.p.  X  $50  X  0.12  =  $3000, 

so  that  the  total  annual  cost  is 

KI   =  3000  H-  7500  =  $10,500. 

The  cost  per  h.p.  year  is 

10,500  ^  500  =  $21, 

and  the  cost  per  h.p.h.  is 

$21    -f-   5OOO    =   $O.OO42. 

During  the  first  3000  hours  we  have  an  additional  load  of 
2000  —  500  =  1500  h.p. 

whose  operating  cost  per  year  is 

1500  h.p.  X  3000  hrs.  X  $0.003  =  $13^500, 
and  the  fixed  charges  are 

1500  h.p.  X  $50  X  0.12  =  $9000. 

The  total  cost  of  this  part  of  the  unit  is 

KI"  =  9000  +  13,500  =  $22,500. 

This  is  for  the  1500  h.p.  unit  during  the  first  3000  hour  period. 
But  during  this  period  we  have  in  use  the  500  h.p.  unit  as  well, 

costing 

500  X  0.0042  X  3000  =  6300, 

so  that  the  total  cost  during  this  3000  hour  period  is 
22,500  +  6300  =  $28,800, 

and  since  the  load  is  2000  h.p.,  the  cost  per  h.p.  is 
28,800  -f-  2000  =  $14.40, 

and  since  this  period  is  3000  hrs.,  the  cost  per  hour  during  this 
period  is  14.40  -r-  3000  =  $0.0048  per  h.p.h. 


204  UNIT  COST  DETERMINATION 

Summary.  —  Cost  per  h.p.h.  during  ist  period  $0.0048. 
"       "       "          "      2nd       "      0.0042 
By  substitution  in  the  equations,  we  get 

Kz    =  — h  0.003  =  0.0012  +  0.003  =  $0.0042. 

3000   X  2000 

for  the  second  period,  and 

„   _  20OO  X  3000  +  2OOO  X  20OO  —  500  X  2OOO 
2000  X  3000  (3000  +  2000) 

X  0.12  X  50  +  0.003, 
or 

KZ  —  0.0018  +  0.003  =  $0.0048    per     h.p.h.     for     the 

second  period,  as  before. 

Example  41.  — A  factory  for  the  manufacture  of  a  gas  en- 
gine costs  $4,000,000  on  which  the  fixed  charges  are  15%. 
The  capacity  of  this  plant  is  600  engines  per  month.  It  sup- 
plies this  number  of  engines  per  month  during  the  first  five 
months,  and  then  300  engines  per  month  for  the  balance  of 
the  year,  due  to  reduced  demand.  Exclusive  of  fixed  charges, 
the  cost  of  manufacture  of  each  engine  is  $150.  Determine  the 
total  cost  per  engine  (a)  during  the  first  period,  and  (b)  during 
the  second  period. 

Solution.  —  If  we  had  a  factory  capacity  of  only  300  engines 
per  month,  its  cost  would  be  $2,000,000  on  which  the  fixed 
charges  would  be  2,000,000  X  0.15  =  $300,000  per  year,  or 
300,000  -T-  12  =  $25,000  per  month;  and  since  300  engines 
are  turned  out  per  month,  this  gives 

2500  -f-  300  =  $83.33  Per  engine, 
wherefore,  for  the  slack  period,  the  total  engine  cost  is 
150  4-  83.33  =  $233.33. 

The  second  $2,000,000  is  required  for  the  extra  demand  of 
300  engines  per  month  during  the  first  5  months.  The  fixed 
charges  thereon  are 

2,000,000  X  0.15  =  $300,000, 


TWO-PART  LOAD  205 

which  must  be  paid  during  this  5-month  period,  together  with 
-f%  of  the  other  $300,000  of  fixed  charges  or 
T62  X  300,000  =  $125,000. 
Total  fixed  charges  assignable  to  this  first  period  are 

300,000  +  125,000  =  $425,000 
or  425,000  -T-  5  =  $85,000  per  month. 

During  this  period,  600  engines  per  month  are  manufac- 
tured, wherefore  the  fixed  charges  per  engine  during  this  rush 

period  are 

85,000  -t-  600  =  $183.33, 

and  the  total  cost  per  engine  during  the  maximum  demand 
period  is 

183-33  +  150  =  $333-33, 
or  $100  per  engine  more  than  during  the  normal  demand  period. 

If  these  engines  are  sold  for  $400  per  engine,  the  net  profit 
per  month  during  the  rush  period  is 

(400  -  333.33)  X  600  =  $40,000, 
and  during  the  normal  period  it  is 

(400  -  233.33)  X  300  =  $50,000. 

It  is  no  doubt  surprising  that  the  net  profits  should  be  less 
during  the  rush  period  than  during  the  normal  demand  period, 
but  this  is  due  to  the  large  investment  to  accommodate  this 
peak  period.  In  this  case  it  would  be  far  better  to  reduce 
the  size  of  the  plant  so  that  it  could  meet  the  total  annual  de- 
mand by  producing  engines  at  a  uniform  rate  throughout  the 
year.  This  would  necessitate  the  tying  up  of  capital  on  prod- 
uct finished  ahead  of  the  demand.  The  interest  thereon 
must  be  charged  against  these  off-season  engines  plus  the  cost 
of  storage,  but  this  would  hardly  amount  to  more  than  $5 
extra  per  engine  as  compared  with  $100  per  engine  extra  on 
the  first  3000  engines  per  year. 

130.  The  application  to  railway  transportation  problems  is 
exactly  similar  as  illustrated  in  the  following  problem: 

Example  42.  —  A  branch  railway  60  miles  long  costs 
$1,800,000  for  ways  and  structures.  Freight  cars  cost  $3500 


2o6  UNIT  COST  DETERMINATION 

each  and  locomotives,  one  for  each  20  cars,  cost  $30,000  each. 
Freight  cars  empty  weigh  5  tons  each  and  carry  30  tons  net 
weight.     Locomotives  weigh  30  tons.     Cost  per  ton-mile  for 
freight  haul  is  0.4  cent.     The  freight   load   amounts  to  120 
cars  per  day  during  July,  August,  and  September,  and  to  20 
cars  per  day  during  the  balance  of  the  year.     Besides  this, 
passenger  service  is  rendered.     Passenger  cars  weigh  7  tons, 
costing  $5000  each,  and  carry,  on  the  average,  40  passengers. 
Cost  per  ton-mile  of  passenger  service  is  0.6  cent.     The  pas- 
senger load  amounts  to  4  cars  per  day  during  June,  July,  and 
August,  and  2  cars  per  day  during  the  balance  of  the  year. 
Determine  the  total  cost  per  ton-mile  during  rush  and  normal 
demand  periods  for  passenger  service,  also  the  cost  per  pas- 
senger-mile if  the  fixed  charges  on  ways  and  structures  are 
10%,  cars  14%,  and  locomotives  14%. 

Solution:    The   first  problem  is  one  of  cost  segregation  as 
between  passenger  and  freight  service.     The  proper  basis  is 
evidently  the  ratio  of  the  maximum  demand.     Thus  the  maxi- 
mum demand  of  passenger  service  consists  of 

4  passenger  cars  @  7  tons  each  ......        28  tons 

4  X  40  X  i5o#  =  24,000  (wt.  of  passengers)       12     " 
and     i  locomotive  @  30  tons  .........  '      30    " 

Totalweight  ............        70    " 

for  2  X  60  =  1  20  miles  per  day  or  70  X  1  20  =  8400  ton-miles 
per  day. 

For  the  freight  service,  the  maximum  demand  consists  of: 
1  20  cars  per  day  @  35  tons  (one  way)  .....       4200  tons 

60  empties  returned  for  loads  assumed  @  5  tons  (one 

way)  ................         300    " 

6  locomotives  ©30  tons  (two  ways)  ......         180    " 

so  that  (4200  +  300)  X  60  =  270,000, 

180  X  1  20  =     21,600 
or  291,600  ton-miles  per  day. 

Of  the  $1,800,000  for  ways  and   structures 
4°o 


291,600 


$I?8oo,ooo  =  $52,000   approximately 


TWO-PART   LOAD  207 

is  to  be  charged  to  passenger  service,  and  the  balance,  or 
1,800,000  —  52,000  =  $1,748,000,  to  freight  service. 

Of  structures  it  should  be  understood  that  certain  items  must 
be  charged  to  the  particular  service  at  once  instead  of  being 
apportioned.  Freight  sheds  and  yards  should  be  charged  to 
freight  service  and  passenger  depots  to  passenger  service 
directly,  and  so  forth. 

The  segregation  of  investments  is  then  as  follows: 

TABLE  90 
PASSENGER  SERVICE 

Cost  F.  C.  rate  F.  C.  total 

Ways  and  structures $52,000  10%  $5,200 

4  passenger  cars 20,000  14%  2,800 

1  locomotive. 30,000  14%  4,200 

Total $12,200 

FREIGHT  SERVICE 

Cost  F.  C.  rate  F.  C.  total 

Ways  and  structures.  .    $1,748,000  10%  $174,800 

180  cars 630,000  13  %  81,900 

5  locomotives 150,000  14%  21,000 

Total $277,700 

SEGREGATION  OF  PASSENGER  FIXED  CHARGE  COSTS  —  NORMAL  PERIOD 
Ways  and  structures $2600 

2  passenger  cars 1400 

i  locomotive 4200 

Total ,         $8200 

$I2,2OO  —  8200  =  $4000. 

The  normal  load  continues  for  12  months,  so  that  the  fixed 
charges  per  month  are 

8200  -4-  12  =  $683.33. 

The  excess  load  continues  for  3  months  during  which  the 
excess  fixed  charges  are 

4000  -r-  3  =  $1333.33  Per  month. 
Total  fixed  charges  per  month  during  rush  period  are 

J333-33  +  683-33  =  $2016.66 
or  2016.66  -r-  30  =  $67.22  per  day. 


208  UNIT  COST  DETERMINATION 

During  normal  period  it  was  found  to  be  only  $683.33  Per 
month,  or 

683-33  •*-  3°  =  $22.77  P^  day. 


During  the  normal  period,  the  train  weight  is 

Locomotive  .................  30  tons 

2  cars  ......................    10    " 

80  passengers  @  i5o#  ........  _6     " 

Total  ...................   46    " 

or  46  X  120  =  5520  ten-miles  per  day. 

Whence  the  fixed  charge  per  ton-mile  is  for  the  normal  period 

22-77  •*•  5520  =  $0.0041. 
The  total  cost  per  ton-mile  is 

0.0041  +  0.006  =  $0.0101  during  normal  period.     (Ans.)  (a) 
Passenger  miles  per  day  during  normal  period  are 

2  X  40  X  1  20  =  9600. 
Ton-miles  per  passenger-mile  are 

5520  ~  9600  =  0.575, 
whence  the  total  cost  per  passenger-mile  is 

o.oioi  X  0.575  =  $0.00581.     (Ans.)        (b)  . 

During  the  rush  period,  the  train  weight  is 

Locomotive  .................   30  tons 

4  cars  ......................   20     " 

160  passengers  @  150$  .......   i£    " 

Total  ...................  62     " 

or  62  X  120  =  7440  ton-miles  per  day. 

Whence  the  fixed  charge  per  ton-mile  during  the  rush  period  is 

67.22  -f-  7440  =  $0.00903, 
and  the  total  cost  per  Ion-mile  is 

'0.00903  +  0.006  =  $0.01503.     (Ans.)      (c) 

Passenger  miles  per  day  during  the  normal  period  are 
4  X  40  X  120  =  19,200, 


VARIABLE   LOAD  —  TWO   PART  209 

so  that  the  ton-miles  per  passenger-mile  are 
7440  -*•  19,200  =  0.3875, 

and  the  total  cost  per  passenger-mile  during  the  rush  period  is 
0.01503  X  0.3875  =  $0.005824.     (Ans.)     (d) 

In  a  similar  way  the  freight  costs  per  ton-mile  during  each 
period  may  be  determined.  Certain  simplifications,  hardly 
warranted  in  practice,  have  been  made  in  the  above  problem 
in  order  to  illustrate  the  application  of  the  principles  without 
too  much  detail. 

131.  In  former  equations,  it  was  assumed  that  the  size  of 
the  unit  and  the  load  were  equal,  —  i.e.,  the  unit  operated  at 
full  load.  This  is  hardly  ever  the  case.  Suppose  then 

(Li  +  Lz)  =  size  of  load  during  the  rush  period, 
(Mi  +  Mz)  =     "     "  units  for  the  rush  period, 

Nz  =  time  in  hours  (days  or  months)  in  the  rush  period, 
LI  =  size  of  load  during  the  normal  period, 
MI  =     "     "  unit  for  the  normal  period, 
(Ni  —  Nz)  =  time  in  hours  (days  or  months)  in  the  normal 

period, 

C  =  cost  per  unit  of  load, 
p  =  fixed  charges  in  per  cent, 
and  a  =  operating  cost  per  unit  of  load  per  hour. 

Then  evidently  the  total  annual  cost  for  the  M  i  unit  is 

Kn  =  MiCp  +  LidNi      ..........    (74) 

and  the  cost  per  unit  of  load  is 


The  total  cost  of  the  excess  load  during  the  rush  period  is 
Kzi'  =  MzCp  +  L2aN2. 

During  this  same  period,  we  also  operate  the  (Mi)  unit  whose 
cost  for  Nz  hours  is 


„  _      iz 

A  12    =  —  TV—  • 


210  UNIT  COST  DETERMINATION 

whence  the  total  cost  for  the  rush  period  is 

/[ ,Nz\  c         7    N 
#1  /    P        *a  2  +    iaN*' 

and  the  cost  per  unit  of    load  per  hour  (h.p.h.,  kw.h.  or  the 
like)  is 

\sfr  t  f        v 

:  r  +  <* (77) 


during  the  rush  period  as  compared  with 

KZI  =  —  •  — — h  & 
Ni    Li 

for  the  normal  period. 

132.  Limitations  on  the  value  of  (Ni)  must  of  course  be  ob- 
served. If  the  time  is  in  hours,  then  (Ni)  cannot  be  greater  than 
8760;  if  in  days,  then  it  cannot  be  greater  than  365;  and  if  in 
months,  then  (A'l)  cannot  be  greater  than  i 2  for  evident  reasons. 
In  a  great  many  cases  it  is  hardly  sufficient  to  divide  the 
load  into  two  parts.  Electric  service  naturally  divides  itself 
into  three  distinct  periods,  the  day,  peak,  and  night  periods. 
So  the  load  on  a  great  many  undertakings  naturally  divides 
itself  into  three  periods,  the  rush,  normal,  and  slack  periods. 
Under  these  conditions,  we  qan  determine  the  costs  during  each 
period  as  follows: 

Li  =  the  mean  load  during  the  slack  period, 
Mi  —  the  capacity  of  unit  during  the  slack  period, 
NI  —  Nz  =  duration  of  the  slack  period, 
(Li  +  LZ)  =  mean  load  during  the  normal  period, 
(Mi  +  Mz)  =  capacity  of  units  during  the  normal  period, 

Nz  —  A^s  =  duration  of  the  normal  period, 
(Li  +  LZ  -f  £3)  =  mean  load  during  the  rush  period, 
(Mi  +  Mz  +  M^)  =  capacity  of  units  during  the  rush  period, 
(Nz)  =  duration  of  the  rush  period, 
C  =  unit  cost  of  the  undertaking, 
p  =  per  cent  of  fixed  charges, 
a  —  operating  cost  per  unit  of  load  per  hour, 


THREE-PART  LOAD 
then  for  the  unit  Mi,  the  total  annual  cost  is 

Kn  =  M£p  +  L&Ni     .    .    . 
and  the  cost  per  unit  of  load  per  hour  is 


211 


(78) 


(79) 


This  is  for  the  slack  period. 

L 


-NT 


N 

FIG.  41.  —  Diagram  for  a  three-  part  load. 

The  excess  load  during  the  normal  over  the  slack  period  is 
(L2)  with  a  corresponding  difference  in  size  of  unit  of  (M2), 
the  total  annual  cost  of  which  is 

Ku'  =  M2Cp  + 
and  the  cost  per  hour  is 


= 


During  this  period   it  operates  for   (A^  —  Na)   hours,   for 
which  time  the  cost  is 


v"  =        Cp(N,  -  N,) 


to  which  must  be  added  the  cost  of  operating  the  (Mi)  unit 
for  the  same  period,  or 


212  UNIT  COST  DETERMINATION 

giving  a  total  for  this  normal  period  of 


-  N9)  (So) 


Dividing  through   by  the  load   (Li  -f  L2)    and  the   hours 
^2  —  Na)  gives  the  cost  per  unit  of  load  per  hour,  or 

^T-  +  «  •    (81) 


which  is  identical  with  equation  (77). 

For  the  rush  period  the  total  cost  of  operating  the  excess 
unit  (Ms)  is 

Kb'  =  M*Cp  +  L3aNz. 

The  cost  of  operating  the  units  (Af2  +  MI)  for  Ns  hours  is 
obtained  by  multiplying  equation  (77)  by  the  load  ( LI  +  L2) 
and  the  hours  N3,  or 

„  „       /Mi       Jl 

-tvis      .=   I  ~^r   +  - 


The  total  cost  of  operation  during  the  rush  period  is  the 
sum  of  KM  and  K^",  or 

v  -1 

(82) 

so  that  the  cost  per  hour,  ..obtained  by  dividing  equation  (82) 
~by  (Li  +  L2  +  Lz)Nz  is 

/Mx       M2       M3\  Cp  (,} 

A23    =    I  — r  +  -T7-  +  -rr-  17-= ; ; r-r-  +  a    .    (d?,) 


If  we  had  (If)  divisions  in  our  load  instead  of  only  three  as 
above,  then  the  costs  during  the  nth  period  can  be  written  at 
once  by  symmetry,  getting 


(84) 


N-PART  LOAD  213 

total  cost  during  the  period,  and  therefore 

_1,M1  M.\  Cp 

-~ 


+  a   .    .    ..............   (85) 

133.  We  wish  to  call  particular  attention  to  the  term 


/Mi  4-  M* 

\Ni  +  N, 


4         i- 

"*  N 


which  we  shall  call  the  service  modulus,  and  denote  by  (5), 
so  that 

c       Ml 

Sl  '-"  If? 


.       5.  =       +       +  ...  +;...  (86) 

We  shall  also  denote  by  Ln'  the  total  load  during  any  period, 
so  that 

Ln'    =    L,    +    L2  +     -      '      -    +    Lnj 

whence  we  may  rewrite  equations  (79)  to  (85)  above  as  follows: 

Kzl  =  S^+  a,      ........    (87) 

L\ 

Ka=StQ  +  a,     ........   (88) 

Li 

KK  =  S,  Q  +  a,  .........  (89) 


and  in  general 

K,n  =  Sn        +  a (90) 


214  UNIT  COST  DETERMINATION 

134.  If  we  plot  the  time  (AO  vertically  and  (5"),  the  service 
modulus,  horizontally  as  in  Fig.  42,  then  the  area  is  the  size  of 
the  unit  (M)  required  for  the  period.  Thus 

~       M, 

Sl  -~-  T; 
so  that  SiNi  =  Mi,  etc.,  the  size  unit  required. 

In  determining  the  costs  of  an  undertaking,  the  work  is 
greatly  simplified  by  first  determining  the  service  modulus 

when  the  costs  for  each 

N  period    may   be   written 

down  at  once. 

Example  43.  —  A  cer- 
tain plant  carries  a  load 
as  follows:  first  period, 
4000  hours  per  year,  mean 
load  400  kw.,  maximum 
500  kw.;  second  period, 
2000  hours,  mean  load 

800  kw.,  maximum  1000  kw.;  third  period,  maximum  2000 
kw.,  mean  load  2000  kw.,  time  2000  hrs.  First  cost  per 
kw.  $70,  fixed  charges  10%,  and  operating  cost  35  cents  per 
kw.h.  Determine  the  cost  per  kw.h.  during  each  period. 

Solution:  In  this  case  the  service  modulus  (S3)  for  the  third 
period,  where 

Mi  =  500,  Ni  =  8000, 

MZ  =  1000  —  500,  NZ  =  4000, 

Ms   =   2OOO   —   IOOO,  NS   =    2OOO, 

Z3    =  2000  kw.h., 
is  equal  to 

c     5OO     IOOO  —  5OO    2OOO  —  IOOO 

^3    —   o I I J 

8OOO  4OOO  2OOO 

or  S3  =  0.0625  +  0-125  +  °-5  =  0.6875. 

Whence  the  cost  per  kw.h.  during  this  period  is 


FIG.  42. 


#23 
#23 


0.6875  X 


70  X  o.io 


2OOO 


0.0035, 


0.0024  H~  °-°°35  =  $0.0059  per  kw.h.     (Ans.) 


SERVICE   MODULUS  215 

Similarly  the  costs  during  the  first  and  second  periods  may  be 
determined. 

135.   In  general,  we  had  for  the  service  modulus, 


_   Mi     .Mt     ,  ,     Mn-i     .     M_ 

~  Ni    +  N2    "  ^  Nn-i    "*"  Nn 


for  (n)  periods. 

For  the  (n  +  i)st  period,  it  is 

_   Mi          Ml  Mn-l     .     Mn     ,     Mn+i 

kj*+i  —  TT     h  "77     r  *    '    *  ~r  ,7         ~r  AT    ~r  Ar      > 

^i       N2  Nn-i       Nn      Nn+i 

the  difference  between  the  two  being 

o,        _  c    ._  Mn+i 

^  n+l  ^n   —    "77  ' 

Nn+1 

When  the  number  of  periods  is  indefinitely  increased,  then 
this  difference  becomes  indefinitely  small  and  we  may  write 

Sn+l    —   Sn    =   dS. 

At  the  same  time  Mn+i  becomes  very  small,  so  that  we  may 

write 

Mn+1  =  dM, 

dM 
whence  M    =  —  ........    (91) 


and  therefore  S  =  +  B  .    .         .    .   (92) 


when  B  =  a  constant  of  integration. 

The  determination  of  the  constant  of  integration  may  offer 
some  difficulty.  Since  the  limits  of  integration  are  from  8760 
hours  to  o  hours,  we  must  use  one  of  these  limits  in  determin- 
ing this  constant.  This  must  be  the  former.  For  we  do  not 
know  the  value  of  (S)  when  N  =  o,  but  when  N  =  8760,  then 

S  =  Si  =  — l  =  — -,  where  (Mi)  is  the  load  when  N  =  8760. 
Ni      8760 

So  then  B  =  — ^-,  and  our  formula  becomes 
8760 

CdM       M\  /    x 

5  =  j  -T+^o (93) 


2l6 


UNIT  COST  DETERMINATION 


The  integration  is  really  carried  out  from  right  to  left  instead 
of  the  usual  order. 

At  the  same  time  the  difference  between  the  maximum  de- 
mand (M)  during  a  given  period  and  the  mean  load  vanishes, 
so  that 

L  =  M 
and  the  cost  per  unit  of  load  per  hour  becomes 


a  . 


(94) 


136.  If  we  plot  the  hours  of  operation  horizontally  as  in 
Fig.  44A  and  the  load  (M)  vertically,  then  we  get  a  load-hour 
curve  as  OO'.  On  the~other  hand,  if  we  plot  (N)  vertically 

N 


FIG.  43- 

as  before  and  the  service  modulus  (S)  horizontally  as  in  Fig. 
43,  then  the  area  under  the  curve  thus  obtained  equals  (M). 
This  we  can  prove  as  follows.  The  area  (^4)  under  the  curve  is 

dA  =  NdS. 


But 

or 

so  that 

and 


dS-dM 

'  W 

dM  =  NdS, 
dA  =  dM, 
A  =  M,  as  stated. 


REPLOT 


217 


The  N-S  diagram  is  of  special  interest  because  it  shows  at 
once  the  load,  hours  of  service,  and  service  modulus. 

Ordinarily  the  load  curve  is  given  as  in  Fig.  44A.  This  is 
the  load  curve  as  it  is  obtained  from  a  graphical  recording  watt- 
meter, the  horizontal  distance  being  the  calendar  time.  But 


730   1460  2190  2920   3650  4380   5110  58.40   6570  7300  8030  8760 
Hours 

FIG.  44.  —  Replot  of  a  typical  power  plant  load. 

to  obtain  even  a  roughly  approximate  equation  between  the 
load  (M)  and  the  time  (N)  would  be  impracticable  for  such  a 
curve  even  if  it  were  not  impossible. 

For  use  in  cost  analysis,  a  replot  is  necessary  in  order  to 
simplify  the  curve.  The  replot  is  made  in  descending  values 
of  (M).  In  order  to  do  this,  first  plot  on  the  ordinate  (OM), 
Fig.  446,  the  point  (Mo)  being  the  maximum  instantaneous 
value  of  the  load  (M)  for  the  year.  Then  draw  a  hori- 
zontal line  as  AB,  in  Fig.  44 A,  at  a  height  (Mi).  Then  the 
number  of  hours  that  the  system  operates  at  a  load  MI  or 
over  is  given  by  the  sum  of  the  intercepts  as  shown,  having 
a  total  value  of  (Ni).  With  coordinates  (Mi)  and  (N^)  we 
can  now  plot  the  point  (P)  in  Fig.  446.  By  shifting  the 
horizontal  line  AB,  Fig.  44A,  we  can  thus  get  as  many  points 


2i8  UNIT  COST  DETERMINATION 

as  we  desire  on  the  replot  of  Fig.  446  and  thus  determine 
this  curve.  The  equation  for  this  is  comparatively  very 
simple,  a  common  algebraic  equation  of  the  third  degree 
usually  giving  a  sufficiently  close  approximation.  In  Fig.  45, 


10       20       30       40       50       60       70       80       90      100     110     120 
FIG.  45.  —  Load  curve  and  replot  of  the  Portland  Central  Heating  Co. 

* 

we  show  the  load  curve  and  replot  of  the  Portland  Central 
Heating  Co.  from  August  to  December  1914. 

137.   Example  44.  —  For  a  certain  power  plant,  the  load 
hour  equation  is  found  to  be 

M  =  12,000  —  o.oooi5./V2. 

The  cost  per  kw.  of  the  installation  is  $70  and  the  fixed  charges 
10%.     If  the    operating  costs   are  0.2  cent  per  kw.h.  deter- 
mine the  total  cost  per  kw.h. 
Solution:  Here  dM  =  — 


and 

so  that 
When 


=  —  0.0003 

S  =  -  o.ooo^N  +  B.  . 
N  =  8760  hrs., 


fNdN 

J  IT' 


(a) 


EXACT   UNIT   COSTS  219 

then  M  =  12,000  —  0.00015  (8760) 2  =  490  kw., 

for  which  S  = 

Substituting  these  values  in  equation  (a)  above,  we  get 
0.056  =  —  0.0003  X  8760  +  B 
B  =  0.056  +  2.628  =  2.684. 

So  that  equation  (a)  becomes 

S  =  2.684  ~~  0.0003 A7", 
and  the  cost  per  kw.h.  is 

v    _  2.684  —  o.ooo$N 


or 


^ 

K2   = 


M 

18.788    -  0.002IN 


70  X  o.io.-f-  0.002. 

+  O.OO2. 


12,000  —  0.00015^2 

We  may  tabulate  values  of  the  cost  per  kw.h.  for  various 
values  of  A^,  as  follows: 

TABLE  91 


N 

Ki 

N 

Kt 

0 

0.0035 

5000 

0.00300 

1000 

o  .  0034 

6000 

o  .  00294 

200O 

0.0033 

7000 

0.00288 

3000 

0.0032 

8000 

0.00283 

4000 

0.0031 

8760 

0.00280 

You  will  note  that  the  cost  of  production  of  power  is  least 
during  the  periods  of  least  load  and  greatest  during  the  periods 
of  greatest  loads.  If  now  the  price  or  rate  is  made  throughout 
proportionate  to  the  cost,  as  it  should  and  must  be,  the  rate 
will  be  lowest  during  the  periods  of  lightest  load,  thus  encourag- 
ing the  filling  of  the  valleys  in  the  power  plant's  load  curve, 
reducing  throughout  thereby  the  cost  and  rate  of  the  power 
service. 


220  UNIT  COST  DETERMINATION 


PROBLEMS 

1.  A  10  h.p.  pumping  plant  runs  at  full  load  for  a  variable  number  of 
hours  per  year.    It  costs  $500  with  10%  of  fixed  charges.    The  cost  of  opera- 
tion is  20  cents  per  hour.    Determine  the  total  annual  and  unit  cost  of 
production  of  the  service.    Plot  the  curves. 

2.  A  25  kw.  electric  lighting  plant  carries  a  load  of  25  kw.  for  1000  hours 
per  year  and  a  load  of  15  kw.  for  an  additional  3000  hours  per  year.    Deter- 
mine the  total  annual  and  unit  cost  of  production  of  the  power,  if  the  plant 
costs  $80  per  kw.  of  capacity,  with  12%  fixed  charges  and  operating  costs 
of  1.2  cents  per  kw.h.  delivered. 

3.  A  power  plant  cost  $60  per  h.p.  installed,  with  fixed  charges  of  12%. 
The  operating  costs  are  0.3  cent  per  h.p.h.    The  plant  carries  a  load  of 
1800  h.p.  for  400  hours  per  year.     During  this  400   hour   period   the 
maximum  demand  is  2200  h.p.    During  the  balance  of  the  year,  the  plant 
carries  a  mean  load  of  600  h.p.  with  a  maximum  demand  of  700  h.p. 
Determine  the  total  annual,  and  the  unit  cost  of  production  of  the  service 
during  each  period. 

4.  A  loo  mile  branch  railroad  is  built  having  a  capacity  of  200  cars  of 
freight  per  day,  and  costing  $2,000,0x30.    The  fixed  charges  are  13%.    The 
cost  of  operation  is  15  cents  per  car-mile  loaded  and  10  cents  per  car-mile 
empty.     During  two  months  in  the  year,  the  system  carries  an  average  of 
1 80  loaded  cars  out,  with  a  maximum  of  200,  and  an  average  return  of  50 
loaded  cars  with  a  maximum  of  60.    During  the  balance  of  the  year,  an 
average  of  80  cars  per  day  is  carried  out  to  the  main  line  loaded,  with  a 
maximum  of  100  and  there  is  an  average  return  of  20  cars  with  a  maxi- 
mum of  30.      Cars  cost  $3000,  locomotives  $20,000,  with  a  capacity  of  20 
cars.     The  weight  of  each  locomotive  is  the  same  as  a  loaded  freight  car. 
Determine  the  total  annual  and  unit  cost  of  service  per  car-mile  during 
each  period. 

6.  A  steam  engine  generating  plant  costs  $75  per  kw.  of  capacity,  with 
10%  fixed  charges.  It  carries  a  mean  load  of  5000  kw.  for  60  hours  per 
year,  with  a  maximum  of  8000  kw.,  a  mean  load  of  3000  kw.  for  1500  hours 
per  year  with  a  maximum  of  4000  kw.  and  a  mean  load  of  1000  kw.  for  the 
balance  of  the  year,  with  a  maximum  of  1500  kw.  The  cost  of  operation 
is  o.i  cent  per  kw.h.  with  a  mean  load  of  5000  kw.,  0.12  cent  per  kw.h. 
when  the  mean  load  is  3000  kw.h.  and  0.14  cent  when  the  mean  load  is 
1000  kw.  Determine  the  total  annual,  and  unit  cost  of  production  of  the 
service  during  each  period. 

6.  What  could  we  afford  to  pay  per  kw.  of  capacity  for  a  turbo-generating 
plant  to  take  the  place  of  the  steam  engine  generating  plant  of  problem  (5), 
if  its  costs  of  operation  are,  for  the  mean  load  of  5000  kw.,  0.12  cent,  for 
the  mean  load  of  3000  kw.,  0.14  cent  and  for  the  mean  load  of  1000  kw., 
0.17  cent,  other  things  being  equal? 


PROBLEMS  221 

7.  A  power  plant  has  an  annual  load-hour  curve  as  given  by  the 
equation 

M  =  5000  -  2oN  4-  0.0024  N* 

and  the  cost  of  operation  per  h.p.h.  is  0.2  cent.  The  first  cost  of  the  plant 
is  $50  per  h.p.  of  capacity  installed,  with  12.5%  fixed  charges.  Determine 
the  total  annual,  average  unit,  and  true  unit  cost  of  production.  Plot  the 
curve  of  the  variation  of  true  unit  cost  with  the  variation  of  the  load. 

8.  A  power  plant  carries  a  load  as  given  by  the  equation 

M  =  50,000  —  5./V. 

The  first  cost  of  the  plant  is  $40  per  h.p.  of  capacity,  installed,  with  10% 
fixed  charges.  The  cost  of  operation  per  h.p.h.  varies  according  to  the 
equation 

A  =  0.3  —  0.00004  M. 
Determine 

(a)  The  total  annual  cost  of  production  per  h.p. 

(b)  The  average  cost  of  production  per  h.p.h. 

(c)  The  true  cost  of  production  per  h.p.h. 

(d)  Plot  the  curve  of  the  variation  of  the  true  cost  of  production  of  the 
service  per  h.p.h.  with  the  variation  of  the  load. 


CHAPTER  VII 

DETERMINATION  OF  SIZE  OF  SYSTEM  FOR  BEST 
FINANCIAL  EFFICIENCY 

Variable  Operating  Cost.  Total  Production  Cost.  Determination  of 
Equation  of  Actual  Load  Curve.  Analysis  of  Heat  Transmission. 
Point  of  Best  Financial  Efficiency.  Determination  of  Size  of  System 
for  Best  Financial  Efficiency. 

138.  In  the  past  we  have  assumed  that  the  operating  cost 
per  h.p.h.  was  a  constant.  This  is  far  from  true,  especially 
in  smaller  sized  plants  and  this  consideration  must  modify 
our  conclusions  considerably  where  best  economy  is  aimed 
at.  So  also  within  any  large  plant,  the  unit  operating  cost 
at  full  load  will  be  considerably  less  than  at  fractional  loads. 

If  then  we  call  ai,  #2,  #3,  etc.,  the  operating  costs,  during 
given  periods  and  corresponding  loads,  we  can  determine  our 
costs  as  before,  but  modified  by  this  variation  in  operating 
cost. 

For  the  first  period,  then,  the  total  annual  cost  Kn  is 

Ku  =  Micp  + 
and  the  cost  per  h.p.h.  is 


i       i 
During  the  second  period,  the  total  cost  for  the  excess  load  is 


where  0%  is  the  excess  operating  cost  per  h.p.h.  required  over 
that  of  the  first  period. 

The  cost  of  operating  the  unit  LI  for  the  Nz  hours  is 

=        cp  N2  + 


222 


VARIABLE  OPERATING  COST  223 

so  that  the  total  cost  of  the  second  period  is  the  sum  of  Ki2' 
and  Kn  ",  or 

cp  +  (L,a,  +  Z,2a2)tf2   .   (95) 


If  we  now   call  a  the  mean  operating  cost  for  the  load 
(Li  +  Lz)  then 

(Lidi  +  L2a2)  =  (Li  -f  1,2)0, 


and          #12  =  N*  +  M2#  +  (Li  +  La) 

The  cost  per  h.p.h.  during  this  period  is  then 


So  in  general 

-^2n    =    -5"n  *   -J-f-  H~  ^nj 
-^  n 

where       Ln'  =  LI  +  L^  +  •    •    •  +  Ln, 
and  the  service  modulus 

«  Ml     ,     ^2     ,  ,     Mn 

^w~^1"i"]V2H  h^Vn' 

And  so,  also,  when  the  length  of  the  period  becomes  in 
finitesmal,  then 


where  5=1  —  —  as  before,  and  (a)  is  variable. 

Knowing  then  M  =  f(N)  ,  and  a  =  F(Af  )  either  by  exact  or 
approximate  equation,  a  complete  solution  of  the  costs  can  be 
made,  either  of  an  existing  system,  or  in  advance  of  its 
construction. 

139.  One  of  the  advantages  in  using  instantaneous  costs 
per  h.p.h.  as  given  by  the  equation 


224     SIZE  OF  SYSTEM  FOR  BEST  FINANCIAL  EFFICIENCY 

is  that  the  conditions  for  minimum  costs  can  be  determined 
by  the  usual  method  of  equating  the  first  differential  to  zero, 
and  the  system  designed  to  meet  these  conditions. 

If  in  the  cost  determination,  the  entire  system  is  treated  as 
a  unit,  so  that  the  costs  include  the  transmission  and  distribu- 
tion as  well  as  plant  costs,  then  the  above  determination  for 
minimum  costs  will  determine  both  the  size  of  plants  and  the 
corresponding  areas  of  distribution  for  any  given  set  of  con- 
ditions, such  that  maximum  financial  efficiency  will  be  attained, 
i.e.,  so  that  the  service  will  be  rendered  at  least  cost. 

In  determining  the  best  size  of  a  system  for  minimum  cost, 
we  must  consider  the  unit  first  cost  (c)  of  the  system  as  vari- 
able, since  this  varies  with  its  size.  Once  this  size  is  deter- 
mined, (c)  becomes  fixed  and  has  a  definite  constant  value. 

Since  the  instantaneous  production  cost  per  unit  of  size 
(h.p.)  per  hour  is 


the  cost  per  hour  for  the  size  (M)  will  be 

K2'  =  Scp  +  Ma, 
and  the  total  production  cost  per  year  will  be 

/87<5o  /"876o 

ScpdN  +  I        MadN  .   .   .    (96) 

The  condition  for  minimum  total  production  cost  is  then 

dK  _ 
dM  = 

The  operating  cost  varies  not  only  with  the  size  of  the  units 
but  also  with  the  per  cent-  of  full  load  that  they  are  carrying, 
because  their  efficiencies  vary  with  fractional  loads  often  very 
greatly.  The  handling  of  the  entire  subject  is  quite  complex 
but  this  should  give  peculiar  satisfaction  to  engineers  for  it 
insures  that  in  future  the  matter  must  be  taken  out  of  the 
hands  of  amateurs. 

Example  45.  —  A  power  plant  carries  a  load  as  shown  in 
Fig.  46.  If  the  cost  of  the  plant  per  h.p.  is  $50,  fixed  charges 


MINIMUM   PRODUCTION   COST 


225 


are    12%,    and    the    operating   cost   per   h.p.h.    is   0.8   cent, 
determine . 

(a)  The  cost  per  h.p.h. 

(ft)  The  total  annual  cost. 

Solution:  The  first  step  necessary  is  to  determine  the  equa- 
tion of  the  curve  shown  in  Fig.  i.  In  order  to  do  this,  we  as- 
sume the  equation  to  be 

M  =  M0  +  AN  +  BN2  +  CN*  +  DN*  .  .   .    .  (i) 

The  longer  the  equation,  i.e.,  the  more  constants,  the 
more  nearly  we  can  approximate  to  the  true  curve.  For  each 


6000 


5000 


4000 


3000 


2000 


1000 


0    1000   2000   3000   4000   5000   6000   7000   8000  8760 

Hours 

FIG.  46.  —  Load  curve  for  example  45. 

constant  in  the  equation  we  can  make  the  approximate  curve 
run  through  one  point  on  the  true  curve.  Thus  in  the  above 
equation  a  there  are  five  constants,  and  we  can  therefore  make 
the  approximate  curve  run  through  five  points  on  the  true 
curve  given.  If  we  had  another  term  (EN5)  then  we  could 
make  it  run  through  a  sixth  point  and  therefore  approximate 
to  the  true  curve  just  so  much  more  closely. 

If  we  express  (M)  in  thousands  of  kilowatts  and  (N)  in  thou- 
sands of  hours,  then  we  can  choose  the  following  five  points 


226     SIZE  OF  SYSTEM  FOR  BEST  FINANCIAL  EFFICIENCY 

of  the  true   curve   through  which  to  make   the   approximate 
curve  as  given  by  equation  i  run.     Thus,  when 

N  =  o,  M  =  5, 
#  =  2,  M  =  4, 
N  =  4,  M  =  2, 
N  =  6,  M  =  i, 
AT  =  8,  M  =  0.9. 

Substituting  in  equation  i,  the  value  N  =  o  and  If  =  5, 
we  get  M0  —  5,  so  the  equation  becomes 

M0  =  5  +  AN  +  BN*  +  CAT3  +  IW4.    .    .   (2) 

Our  task  is  now  to  determine  the  values  oi  A,  B,  C,  and  D. 
We  shall  follow  this  through  in  considerable  detail. 

Substitute  in  equation  2  the  value  N  =  2  and  M  =  4,  and 
we  get          4  =  5  +  2^+  4^+  8C  +  i6D, 
or  o  =  i  +  2  A  +  4B  +  8C  +  i6Z)     ....      (3) 

Then  substitute  in  equation  2  the  values  Ar  =  4  and  M  =  2, 
and  we  get 

2  =*=  5  +  4,4  +  i65  +  64C  +  256^ 
or  o  =  3  +  4A  +  i6B  +  6^C  +  256!)   ...      (4) 

The  substitution  in  equation  2  of  the  values  N  =  6,  M  =  i 
gives  o  =  4  +  6^4  +  36^  +  2i6C  +  1296!), 

or  o  =  2  +3^  +i85  +  io8C  +  648!).    .    .      (5) 

And  finally  the  substitution  of  the  values  N  =  8,  M  =  0.9 
gives  o  =  4.1  +  8^4  +  64$  +  5i2C  +  4096,0  .  (6) 

We  have  thus  the  four  following  equations: 

o  =  i  +  2A  +  4B  +8C  +  i6D (3) 

o  =  3  +  ^A  +  i6B  +  64C  +  2$6D   .    .    .     (4) 

o  =  4  +  6A  +  36,8  +  2i6C  +  i296D   .    .      (5) 

and  o  =  4.1  +  84  +  64.8  -f-  5i2C  +  4096^)    .      (6) 

We  can  now  eliminate  (A)  between  equations  3  and  4  by  mul- 
tiplying equation  3  by  2  and  subtracting  from  equation  4, 
thus  0  =  2+4^+  SB  +  i6C  +  32Z> 

o  =  3  +  4A  +  i6B  +  64C  +  256!)  ...      (4) 
subtracting,  o  =  i  +    SB  -f  486'  +  224!)  ...      (7) 


LOAD-HOUR   CURVE  227 

We  can  also  eliminate  (A)  between  equations  3  and  5  by 
multiplying  the  former  by  3  and  subtracting  from  the  latter, 
thus 

o  =  3  +  64  +  i2B  +  24C    +  48£> 
o  =  4  +  6A  +  36^  +  2i6C  +  1296!) 
subtract        o  =  i  +  24$  +  iQ2C  +  12487)  .    .      (8) 

And  finally,  we  can  get  a  third  equation  with  (A)  eliminated, 
by  multiplying  equation  4  by  2  and  subtracting  from  equation 
6,  thus          0  =  6          +  SA  +  32^    +  I28C  +  512!) 
o  =  4.1       +  8.4  +  64^    +  5i2C  +  4096!) 
subtract        o  =  --  1.9  +  32$  +  384C  +  3584!)  (9) 

This  gives  us  the  three  following  equations: 

o  =  i  +  SB  +  48C  +  2240  ......  (7) 

0  =  1  +  24^  +  iQ2C  +  1248!)  .....  (8) 

o  =  --  1.9  +  32#  +  384C  +  3584,0.    .    .  (9) 

We  now  proceed  to  eliminate  (B)  from  these  equations. 
Multiply  equation  7  by  3  and  subtract  from  equation  8.  This 
gives  0  =  3  +  24$  +  i44C  +  672!) 

0=1    +   24#  +   IQ2C  +   I24&D 


subtract        o  =  --  2  +    48C  +  576^  .....      (10) 

Also  multiply  equation  7  by  4  and  subtract  from  equation  9. 
This  gives     o  =  4  +  32$  +  I92C  +    Sq6D 

o  =  --  1.9  +  32^  +  384C  +  3584^ 
subtract        o  =    -  5.9  +  1926'  +  2688£>    .    .      (n) 

We  have  now  only  the  two  equations  following,  with  only 
two  unknowns,  C  and  D. 

o  =  -       2  +  48C  +    576!)       .....      (10) 
o  =  -  5.9  +  i92C  +  2688Z). 

Multiply  equation  10  by  4  and  subtract  from  equation  n. 
This  gives     o  =  —     8  +  I92C  +  2304!) 
o  =  --  5.9  -f  !92C  +  2688D 
subtract        o  =  --  2.1  +    384!) 


-r 


So  that         D  =  --  —  =  —  0.00547 


228     SIZE  OF  SYSTEM  FOR  BEST  FINANCIAL  EFFICIENCY 
By  substituting  this  value  of  D  in  equation  10,  we  get 

o  =  -  2  +  48C  -  ±±  X  576, 
3°4 

or  o  =  -  2  +  48C  -  3.15. 

So  that        C  =  -S^S  =  0.1073. 
48 

We  have  now  the  values  of  both  C  and  D.     Substituting 
these  in  equation  7  gives 


48  384 

or  o  =  i  +  SB  +  5.15  —  1.225. 

So  that         o  =  SB  +  4.925, 


and  B  =  -  =^p    =  -  0.6156. 

8 

Dividing  equation  3  by  2  gives 

o  =  0.5  +  A  +  2B  +  4C  +  8J9, 

and  substituting  the  values  found  for  B,  C,  and  D  gives 

o  =  0.5  +  -4  ••  1-23125  +  .429167  -  .04375, 
or  o  =  0.929  —  1.275^4, 

so  that         A  =  0.346. 

Substituting  these  values  in  equation  2,  we  get 
M  =  5  +  0.346A7"  —  0.6156^  +  0.1073^  —  0.00547^  (12) 

as  the  approximate  equation  of  the  curve.  The  crosses  (x)  in 
Fig.  46  indicate  points  calculated  from  this  equation.  The 
conformation  is  very  close. 

We  can  now  determine  the  service  modulus  (5),  since  by 
equation  12 

dM  =  (0.346  -  1.2312^  +  0.3219^2  -  o.o2iSSN3)dN, 
„        .    '^dM 

O    —     I  ~rr~j 


N 

'•76°  (0.346  -  1.2312^  +  0.3219^  -  o.Q2i88#3)  dN, 

N 


LOAD-HOUR   CURVE  229 

or  5  =  0.346  logeAr-i.23i2Ar-f-o.i6ioAr2-o.oo729/V'3-f-C.1  (13) 
Where  C'  is  a  constant  of  integration.  By  equation  12,  when 
N  =  8.760,  M  =  0.71,  so  that  when  N  =  8.760 

S  =  jl±  =  0.08105. 
8.760 

Substituting  this  value  of  (S)  and  of  (N)  in  equation  13, 
gives 

0.08105  =  0.346  X  2.1702  —  1.2312  X  8.76  +  0.161 

X  (8.76)2  -  0.00729  X  (8.76)3  +  C1, 
so  that  C'  =  2.6612 

and 
5  =  o.3461oge^V  —  i.  2$i2N+o.i6iN2  —  0.00729^  +  2.6612  (14) 

The  cost  per  kw.h.,  is 

*  =  ir  +  0 

M 

S  S 

or  K  =  50  X  o.i2—  -  +  0.008=  6  —  +  0.008. 

M  M 

Substituting  the  values  of  (S)  and  (M)  found,  we  get 
„  _  6(0.346  logeN  —  1.2312^  +  o.  i6iN2  —  0.00729^  +  2.6612) 


1000(5  +  0.346^  —  0.6156^  +  o.iojsN3  —  o.  00547^ 
+  0.008  ................      (15) 

as  the  cost  per  kw.h.  in  terms  of  N.     (Ans.  a.) 
The  total  cost  per  hour  for  the  load  (M)  is 

Kl  =  6ooo5  -f  o.ooSM  X  iooo2  =  6oooS  +  SoooM, 

where  (M)  is  expressed  in  thousands  of  kilowatts  and  N  in 
thousands  of  hours. 

The  total  production  cost  per  year  is 

/8.7<5o  /*8.76o 

MK'dN  =  6000   /        SdN 
Jo 

/.76o 
MdN  .    .     (16) 


230    SIZE  OF  SYSTEM  FOR  BEST  FINANCIAL  EFFICIENCY 
The  integral 

f*«-ff%«-f%x*  =/'"«.(„, 


=  5000, 

dM 

since  dS  =  -7- 

N 

The  total  fixed  charges  per  year  are 

6    /  S  dN  =  6000  X  5000  =  $30,000. 

The  number  of  millions  of  kilowatt-hours  (thousands  of  kilo- 
watts multiplied  by  thousands  of  hours)  produced  per  year  is 
given  by 


=   r 


+  0.346^  -  0.6156^2  +  0.1073^  _  o. 


=  5  X  8.76  —  0.173  X  76.7376  —  0.2052  X  672.2214     .  . 

'+  0.02682  X  5888.6595  —  0.001094  X  51584.66 
=  43.80  -  13,276  -  137.94  +  157-934  ~  56.434 
=  20.636  millions  of  kilowatt-hours,  costing 
=  20.636  X  8000  =  $165,088 

whence  the  total  annual  cost  is 

30,000  +  165,088  =  $195,088.     (Ans.  ft.) 

Example  46.  —  If  in  the  above  example  the  operating  cost 
per  thousand  kw.h.  is  given  by  the  equation 

a  =  (10  -  M) 

between  500  and  5000  kw.,  where  M  is  again  expressed  in 
thousands  of  kw., 

(a)  Determine  the  cost  per  kw.h.  and 

(b)  Determine  the  total  annual  cost. 

Solution:  In  the  previous  problem,  the  operating  cost  (A) 
was  assumed  constant.     In  this  problem,  it  is  assumed  variable 


TOTAL   PRODUCTION  231 

in  accordance  with  the  above  equation.  It  is  evident  that  the 
value  of  (S)  ,  the  service  modulus,  will  not  be  affected  by  varia- 
tions in  the  operating  cost.  Equation  15  above  will  not  be 
changed  except  in  the  last  term,  which  must  be  changed  to 
the  new  value.  We  therefore  get,  for  the  cost  per  kw.h.  (  K) 
K_  6(0.346  \ogeN  —  1.2312^  +  o.i6i7V2—  0.00729^  -j-  2.6612) 
1000(5  +  0.346^-  o.6i56N2  +  0.1073^  ~  0.00547^) 

(10  —  M)      /A          v 
+  i  -  '-      (Ans.  a), 
1000 

where  as  before 

M  =  5  +  0.346^  —  0.6156^  -f  0.1073^*  —  o.ooStfN*. 
The  total  cost  per  year  is  then 

/.76o  _         /*8.7<x> 

SdN  +  iooo2   I         aMdN. 
»/o 
/8.76o° 
S  dN  =  $30,000  as  before 

/8.y6o  r  8.760 

aMdN  =  iooo   /         (10  -  M)M  dN 
Jo 

/8.?6o 
(loM  -  M2)dN. 

Substituting  for  M  its  value  in  terms  of  (N)  as  given  above, 
we  get 

1000^(25  -  o.i2AT2  +  0.426N3  -  o.4527Ar4  +  0.1359^ 
—  o.  01825  N6  +  0.001174^  —  o.oooo3Ars)^A7' 

=  iooo(25Ar  —  0.04^  +  o.io65Ar4  —  o. 


i 
=  ioooAr(25  —  O.O4A7"2  +  0.1065^  —  o.o9O54A^4  +  O.O3365M 

—  0.00261^  +  0.000147  N7  —  0.0000033  A'8),,8-76 
=  8760(25  -  3.07  +  71.6  -  533.0  +  1166.1  -  1177-1 

+  580.7  -  115.4) 

=  8760  X  14.2  =  $124,392  operating  costs, 
so  that  the  total  annual  production  costs  are 

Ki  =  30,000  +  124,392  =  $154,392.     (Ans.  6.) 


23  2    SIZE  OF  SYSTEM  FOR  BEST  FINANCIAL  EFFICIENCY 

While,  as  may  be  seen  from  the  above,  the  work  of  solving 
such  cost  problems  requires  skill  and  considerable  painstaking 
effort,  it  certainly  permits  of  real  and  conclusive  deductions. 

140.  In  the  following  problem  of  determining  the  size  of  a 
heating  system  for  best  financial  efficiency,  we  have  assumed 
the  heat  to  be  transmitted  in  water  instead  of  steam.  It  may 
be  of  interest  to  note  the  comparative  cost  of  transmitting 
heat  energy  by  steam  and  by  water. 

We  can  allow  approximately  a  velocity  of  100  ft.  per  second 
for  steam  and  5  feet  per  second  for  water,  the  velocity  of  steam 
being  twenty  times  as  great  as  that  of  water.  The  heat  capacity 
of  water  is  one  B.t.u.  per  pound,  or  62.5  B.t.u.  per  cu.  ft.  The 
heat  capacity  of  steam  varies  with  the  pressure  (density)  in 
accordance  with  the  following  table. 

TABLE  92 
HEAT  CAPACITY  FOR  STEAM  ABOVE  150°  F. 


Pressure, 

Temp., 

Heat 

Heat  per  cu.  ft. 

Heat  per  cu.  ft. 

Ibs.  abs. 

dry  sat. 

per  cu.  ft. 

per  degree  F. 

per  degree  X  20 

10 

193 

26.5 

0.62  B.t.u. 

12.4 

15 

213 

39-o 

0.62 

12.4 

2O 

228 

5i-5 

0.66 

13.2 

SO 

281 

124 

0-95 

19.0 

IOO 

328 

240 

i.  35 

27.0 

150 

358 

357 

1.71 

34-2 

250 

401 

583 

2.32 

46.4 

350 

432 

814 

2.90 

58.0 

The  third  column  is  obtained  by  dividing  the  total  heat  per 
pound  of  the  steam  above  150°  F.  by  the  number  of  cubic  feet 
per  pound  of  steam,  thus  getting  the  B.t.u.  per  cu.  ft.  The 
fourth  column  is  then  obtained  by  dividing  the  heat  per  cu. 
ft.  by  the  temperature  of  the  steam  above  150°.  But  inas- 
much as  steam  may  be  allowed  to  travel  20  times  as  fast  as 
water,  we  have  multiplied  this  column  by  20,  giving  the  last 
column.  This  shows  12.4  to  58  B.t.u.  for  steam  as  compared 
with  water  at  62.5  B.t.u.  per  cu.  ft.  per  degree.  A  number  of 
important  factors  must  be  borne  in  mind.  If  the  pressure 


TYPE   OF   HEATING   SYSTEM  233 

is  over  ioo#,  then  extra  strong  pipe  and  fittings  must  be  used. 
Since  these  are  far  more  expensive  than  standard,  the  cost  of 
transmission  is  greatly  increased,  which  does  not  permit  of  a 
fair  comparison  with  that  of  hot  water  at  nominal  pressures. 
Therefore  the  real  capacity  of  steam  is  really  only  12.4  to  27 
B.t.u.  as  compared  with  water  at  62.5. 

On  the  other  hand,  while  water  transmission  requires  as 
large  a  return  pipe  as  outgoing  pipe,  that  for  steam  may  be 
very  small,  and  if  the  condensate  is  wasted,  the  return  pipe 
may  be  done  away  with  altogether.  But  in  the  latter  case, 
the  cost  of  purchasing  or  pumping  the  water  thus  wasted  must 
be  added  to  the  total  service  cost. 

But  even  without  a  return  pipe  for  the  steam,  and  allowing 
nothing  for  the  cost  of  water  thus  necessarily  wasted,  we  have 
12.4  to  27  B.t.u.  for  steam  as  compared  with  half  of  62.5,  or 
31.25  B.t.u.,  for  water  transmission.  So  that  even  under  these 
conditions  the  cost  of  steam  transmission  is  at  least  16% 
greater  than  for  water.  The  cost  of  water  wasted  usually 
runs  from  one  to  ten  cents  per  1000  Ibs.,  a  fairly  significant 
item  itself. 

The  conclusion  cannot  be  avoided,  then,  that  in  spite  of  the 
fact  that  most  central  heating  stations  are  designed  for  steam 
transmission,  heat  transmission  by  water  is  far  more  economical. 
If,  in  the  hot  water  system,  instead  of  heating  water  .in  the 
boilers,  we  generate  steam,  using  the  steam  for  the  generation 
of  power,  and  using  the  exhaust  steam  (atmospheric  pressure, 
212°)  for  the  heating  of  the  water,  we  have  an  incomparably 
more  economical  system. 

Example  47.  —  Assume  that  the  cost  per  bo.h.p.  of  a  central 
heating  plant  varies  as  curve  A,  Fig.  47,  and  that  the  cost  of 
transmission  pipe  per  foot  complete  in  place  varies  as  curve 
A,  Fig.  48,  and  that  the  operating  cost  is  one  cent  per 
bo.h.p.  hour,  neglecting  radiation  loss  in  transmission,  deter- 
mine the  area  of  transmission  for  maximum  economy;  in  blocks 
of  200  ft.,  if  the  maximum  demand  is  268,000  B.t.u.  per  block 
per  hour  and  the  mean  annual  load  67,000  B.t.u.  per  block 
per  hour,  fixed  charges  10%. 


234     SIZE  OF  SYSTEM  FOR  BEST  FINANCIAL  EFFICIENCY 


$120 
100 

80 

ffi 
I60 

40 


20 


GOO         700         800         900       1000 
FIG.  47.  —  First  cost  per  horse  power  of  heating  plant  assumed  in  example  47. 


0         .100         200        300        400        500 

Bo.  H.P. 


$2.40 

2.00 
1 
|l.60 

M 

d 

|l.20 

! 

0.80 
-    0.40 

k 

s> 

J,s 

^ 

^^ 

.>* 

^ 

^ 

^ 

s* 

^ 

s* 

^ 

^ 

^ 

^ 

^ 

^ 

^ 

^ 

^ 

Cross-sectional  Area  of  Pipe 


U 

FIG.  48. — Variation  of  cost  of  pipe  with  size  assumed  in  example  47. 


SIZE   OF   HEATING   SYSTEM  235 

Solution:  The  total  production  cost  per  year  is 

Ki  =  Mcp  H-  LaN, 
where    p  =  o.io, 
a  =  o.oi, 
N  =  8760. 

Assuming  the  shape  of  the  area  served  to  be  square,  the 
number  of  blocks  on  each  side  being  (B)  blocks  from  the  plant, 
then  the  total  number  of  blocks  served  will  be  2B(B  +  i), 
in  which  B  is  always  odd.  Since  each  block  requires  268,000 
B.t.u.  maximum,  or  8  bo.h.p.,  the  total  maximum  load  will  be 

M  =  2B  (B  +  i)  X  8  =  i6B  (B  +  i)  bo.h.p., 
and  the  mean  annual  load  (L)  will  be 

L  =  J  X  i6B  (B  +  i)  =  4B  (B  +  i)  bo.h.p. 
whence  the  annual  operating  cost  will  be 
LaN  =  4B  (B  +  i)  X  o.oi  X  8760  =  350.4^  (B  +  i)  dollars. 

We  must  now  determine  (C),  the  cost  of  the  system  per 
bo.h.p.  Curve  A,  Fig.  47,  gives  (Ci)  the  cost  per  bo.h.p. 
of  the  heating  plant.  So  we  must  only  determine  (Cz)j  the 
cost  per  bo.h.p.  of  the  transmission  according  to  the  cost 
data  given  in  Fig.  48. 

Allowing  200°  F.  as  the  maximum  temperature  of  the  out- 
going water  and  150°  as  the  temperature  of  the  return  water 
we  can  carry  50  B.t.u.  per  Ib  or 

50  X  8j  =  417  B.t.u.  per  gallon. 

Since  each  block  requires  268,000  B.t.u.  maximum  per  hour, 
or  4470  B.t.u.  per  minute,  there  will  be  required 

4470  -4-417  =  ii  gallons  per  minute  per  block. 

This  will  require,  for  a  friction  head  of  about  one  foot  per 
hundred  feet  of  pipe,  a  ij"  pipe. 

The  length  and  size  of  distribution  pipe  may  be  computed 
as  follows: 

(1)  For  one  square  block  transmission  area* 

Pipe  required  4  X  i  (/)  of  n  g.p.m.  capacity,  where  I  =  200  ft. 

(2)  For  a  square  of  3  blocks  on  a  side 

Pipe  required  is  4  (/)  (i  +  3)  of  11  g.p.m.  capacity. 
4  (/)  X  2  of  22  g.p.m.  capacity. 


236    SIZE  OF  SYSTEM  FOR  BEST  FINANCIAL  EFFICIENCY 

(3)  For  a  square  of  5  blocks  on  a  side 

Pipe  required  is  4  (/)  (i  +  3  +  5)  of  n  g.p.m.  cap. 
4  (/)  X  4  of  22  g.p.m.  cap., 
4  (0  X  2  of  44  g.p.m.  cap. 

(4)  For  a  square  of  7  blocks  on  a  side 

Pipe  required  is  4  (/)  (i  +  3  +  5  +  7)  of  n  g.p.m.  cap. 

4  (/)  X  6  of  22  g.p.m.  cap, 

4  (I)  X  4  of  44  g.p.m.  cap, 

4  (/)  X  2  of  66  g.p.m.  cap. 

(5)  In  general  for  a  square  B  blocks  on  a  side 

Pipe  required  is  4  (/)  (i  +  3  +  5  +  7  +  9  +  •    •    •  +  B)  ft.  of 
ii  g.p.m.  cap, 

4  (/)  X  (B  -  i)  ft.  of  22  g.p.m.  cap, 
4  (/)  X  (B  -  3)  ft.  of  44  g.p.m.  cap, 
4  (/)  X  (B  —  5)  ft.  of  66  g.p.m.  cap, 
4  (/)  X  (B  -  7)  ft.  of  88  g.p.m.  cap, 

etc. 

The  cost  of  the  i|"  pipe  according  to  Fig.  48  is  34  cents  per 
foot,  and  since  the  cost  in  place  is  assumed  in  direct  propor- 
tion to  the  area  of  the  pipe,  the  cost  per  foot  of  the  22  g.p.m. 

pipe  is 

34  X  2  =  68  cents  per  ft.; 

of  the  44  g.p.m.  pipe  is 

34  X  4  =  $1-36  per  ft,  etc. 
The  total  length  of  i|"  pipe  required  is 

4  X  200  (i  +  3  +  5  +  7  +  •    •    •  +  B)  ft, 

800  X  (B  +     ]  =  200  (B  +  i)2  ft, 

\       2         / 

and  its  cost  is 

200  (B  +  i)2  X  0.34  =  68  (B  +  i)2  dollars. 

The  cost  of  the  22  g.p.m.  pipe  is 

4  X  200  (B  -  i)  X  0.68  =  544  (B  -  i)  dollars; 

of  the  44  g.p.m.  pipe,  it  is 

4  X  200  (B  -  3)  2  X  0.68  =  544  X  2  (B  -  3). 


SIZE   OF   HEATING   SYSTEM  237 

Of  the  66  g.p.m.  pipe,  it  is 

4  X  200  (B  -  5)  3  X  .68  =  544  X  3  (B  -  5), 
etc.,  up  to 

4  X  200  X  2  X 

$7000 

11  Gpm. 


X  -68  =  544  X  2, 


FIG.  49. —  Heat  distribution  layout  for  example  47. 

whence  the  total  transmission  cost  is 

C2  =  68  (B  +  i)2+  544  (B  -i)  +  2  (B  -3)  +  3  (B  -  5) 


+  4  (B  -  7)  +  5  (B  -  9)  + 


Letting 


X   2 


we  get 


C2  =  68  (B  +  i)2  +  5447. 


238     SIZE  OF  SYSTEM  BEST  FOR  FINANCIAL  EFFICIENCY 

Dividing  through  by 

M  =  i6B(B  +  i) 
gives  the  cost  of  transmission  per  bo.h.p.,  or 

_  ,  _  (B  + 1)  34J 


B  B(B  +  i) 

The  variation  of  (J)  and  C2'  with  (B)  is  as  follows: 

B i    3    5    7    9    ii    13    *5    i7    19    21 

/ o    2     8    20    40    70   112   168   240   330   440 

Cz' 8.5         11.3        1.42         17.0        IQ.8          22.6          25.6          28.3         31.2          34.0        36.8 

Assuming  now  that  we  can  represent  the  cost  of  the  heating 
plant  (Ci)  per  bo.h.p.  by  the  following  equation  between  600 
and  1 200  bo.h.p., 

Ci  =  61.66  —  O.O35-M"  +  O.OOOOH33M2. 
We  get,  since 

M  =  i6B(B  +  i), 
that      Ci  =  61.66  -  0.035  X  i6B(B  +  i) 

+  0.00001133  X  16  X  i652(5  +  i)2 
or         Ci  =  61.66  -  0.56^(5  +  i)  +  o.oo2952(5  +  i)2. 
Where  the  total  first  cost  (C)  of  the  system  per  bo.h.p.  is 

/~*     S~**  I          /""*    / 

or          C  =  61.66  -  0.565(5  +  i)  +  o.oo2952(5  +  i)2 


B 
the  annual  fixed  charges  per  bo.h.p.  are 

Cp  =  6.166  -  o.o$6B(B  +  i)  +  o.ooo29^2(5  -f  i) 


5  5(5  +  i) 

Since  the  total  operating  costs  per  year  are 

LaN  =  350.45(5  +  i), 
the  cost  per  installed  bo.h.p. 

LaN  =  350.45(5  4-  i) 
M  i65(5  +  i)      =  2I'9' 


SIZE   OF   HEATING   SYSTEM 

the  total  production  cost  per  bo.h.p.  per  year  is  therefor  . 
K  =  21.9  +  6.166  -  o.o$6B(B  +  i) 


239 


0.425 


or  K  =  28.066  -  o.o$6B(B  +  i)  +  o.ooo29#2(£  +  i)2 


Differentiating  (K)  with  respect  to  (B)  and  equating  to 
zero,  will  give  very  complex  results.  We  can,  however,  make 
a  solution  by  plotting.  Thus 

B  .............    i        3        5        7 


ii    13     15 


17     19 


K 


28.7     29.  i     28.1     27.5     27.4     28    30    32.8    41.2     52    67.8 


The  maximum  financial  efficiency  evidently  takes  place  at 
B  =  9,  an  area  of  service  of  180  blocks.     Under  this  condition 
M  =  i6B(B  +  i)  =  16  X  9  X  10, 
M  =  1440  bo.h.p. 

The  curve  for  the  variation  of  the  production  of  service  cost 
per  bo.h.p.  per  year  with  the  length  of  area  served  is  given  in 


5 


21 
19 
17 
15 

9 

5 

3 
1 
0 

' 

P—  - 

-^z* 

^  ~ 

=T 

*^— 

^.  •• 

,*«— 

s»^> 

t—  — 

*^- 

^  •• 

^  •* 

/ 

/' 

/ 

\ 

\. 

\ 

. 

^ 

^* 

$30 

$40            $50           $60 

Cost  per  Bo.H.P.  Year 
FIG.  50.  —  Variation  of  the  unit  cost  of  service  with  the  size  of  the  system 
according  to  example  47. 


24o     SIZE  OF  SYSTEM  BEST  FOR  FINANCIAL  EFFICIENCY 


Fig.  50.  Note  that  after  the  point  of  best  economy  is  passed 
the  increase  in  cost  is  very  rapid.  The  cost  of  service  of  a 
heating  system  that  was  considerably  too  large  would  be  far 
in  excess  of  the  individual  plant  costs.  However,  the  results 
obtained  above  must  not  be  generalized,  as  they  apply  only 
under  the  conditions  assumed  in  the  problem.  But  under 
these  conditions,  a  steam  heating  system  is  still  more  expen- 
sive and  capable  of  serving  only  a  smaller  area  economically. 
Another  factor  that  must  be  borne  in  mind  is  load  density. 
We  have  in  the  above  problem  assumed  uniform  load  density. 
The  load  density  may  be  very  much  greater  or  less  than  that 
assumed,  while  on  the  other  hand  the  load  density,  instead  of 
being  uniform,  may  vary  over  the  area  in  a  great  many  essen- 
tially different  ways.  The  effect  of  these  variations  on  the 
system's  size  and  economy  afford  study  of  the  most  valuable 
kind  in  practice. 

PROBLEMS 

1.  Assume  the  annual  load  curve  to  be  as  in  Fig.  51,  the  first  cost  of 
the  power  plant  to  be  $60  per  h.p.  installed.,  fixed  charges  12%,  and 
operating  costs  to  be  0.6  cent  per  h.p.h. 


5000 


4000 


^3000 


2000 


1000 


\ 


1DOO       2000       3000       4000       5000      6000       7000       8000    876.0 
Hours 

Fig.  51.  —  Load  curve  for  problem  i. 


PROBLEMS  241 

(a)  Determine  the  total  annual  cost  of  production, 

(b)  Determine  the  cost  of  production  per  h.p.h.  and 

(c)  Plot  the  curve  of  variation  of  cost  per  h.p.h.  with  the  load. 

2.  If  in  problem  (i),  the  operating  cost  in  dollars  per  h.p.h.  is  given  by 
the  equation 

a  =  0.01125  —  0.00000125  M 

and  all  other  things  are  equal,  determine 

(a)  the  total  annual  production  cost, 

(b)  the  number  of  h.p.h.  produced  per  year, 

(c)  the  cost  of  service  per  h.p.h.  and 

(d)  plot  the  curve  of  variation  of  cost  of  service  per  h.p.h.  with  the 
size  of  the  load. 

3.  In  a  power  plant  the  load  varies  with  time  according  to  the  equation 

M  =   5OOO  +  2N  —  O.OO02JV2, 

and  the  cost  of  operation  per  kw.h.  varies  according  to  the  equation 

a  =  0.015  —  o.oooooiAf. 
The  first  cost  of  the  plant  per  kw.h.  is  $75,  fixed  charges  10%.    Determine: 

(a)  the  total  annual  production  cost, 

(b)  '  kw.h.  produced  per  year, 

(c)  "  average  cost  per  kw.h., 

(d)  "  true  cost  per  kw.h., 

(e)  "  maximum  cost  per  kw.h.  and 

(f)  "minimum     "      "    kw.h. 


CHAPTER  VIII 
DETERMINATION  OF  TYPE  AND  SIZE  OF  UNITS 

Stand-by  Units.    Number  of  Units  in  a  Plant  for  Given  Load  for  Best 
Financial  Efficiency.    Economy  of  Units  at  Fractional  Loads. 

141.  In  considering  the  matter  of  the  proper  selection  of 
the  size  of  units  in  any  given  plant,  too  much  emphasis  has  been 
placed  on  the  necessity  of  a  reserve  unit.  Like  most  things,  a 
reserve  unit  is  desirable  in  inverse  proportion  to  its  cost. 
When  this  cost  rises  above  the  probable  gain,  it  is  no  longer 
good  judgment  to  provide  it. 

The  first  question  is,  how  long  will  a  given  piece  of  ma- 
chinery run,  without  a  shutdown,  under  conditions  of  aver- 
age care?  While  this  is  a  very  difficult  matter  to  decide  on, 
yet  as  a  rule  the  number  of  hours  of  run  without  shutdown  will 
be  in  direct  proportion  to  the  life  of  the  equipment.  But 
with  this,  we  must  consider  the  nature  of  the  equipment  as 
well.  For  example",  the  life  of  a  motor  is  twice  that  of  an 
engine,  therefore  we  may  expect  twice  as  long  a  run.  But  as 
a  matter  of  fact,  we  will  get  more  than  that,  because  the 
shock  of  reciprocating  parts  is  absent  in  the  motor.  So  as  a 
whole,  the  period  of  continuous  run  is  far  greater  for  all  rota- 
tive classes  of  machinery,  such  as  generators,  turbines,  centrif- 
ugal pumps  and  the  like,  than  for  the  reciprocating  classes, 
as  engines,  compressors  and  such.  < 

That  shutdown  will  occur  is  certain,  for  no  machine  is 
built  to  run  continuously  during  its  entire  life,  without  certain 
parts  being  either  adjusted  or  repaired,  or  entirely  replaced. 
Among  such  parts  are  bearings,  crosshead  guides,  stuffing  boxes 
and  the  like.  However,  the  period  of  necessary  shutdown 
may  be  anticipated,  and  for  that  matter  usually  must  be  an- 
ticipated to  avoid  severe  damage  to  the  machine,  due  to  break- 

.242 


STAND-BY  UNITS  243 

down.  And  its  repair  may  therefore  be  accommodated  to  the 
load  being  carried. 

But  this  is  not  the  primary  consideration  in  the  matter  of 
dividing  our  load  among  a  number  .of  units.  The  menace  of 
shutdown  is  not  nearly  so  great  as  one  would  be  lead  to  believe 
by  the  emphasis  placed  on  it  by  so  many  authors.  First-class 
machinery,  run  with  average  care,  under  normal  loads,  may 
be  depended  on  to  run  at  least  a  certain  period  without  any 
danger  of  shutdown.  They  will  then  require  certain  repair, 
after  which  they  may  again  be  relied  upon  with  reasonable 
certainty. 

142.  The  reason  for  the  design  and  installation  of  any  system 
is  the  rendering  of  a  given  service.  The  primary  object  for  the 
subdivision  of  the  plant  into  several  units  must  be  to  render 
this  service  at  as  low  a  cost  as  possible.  We  must  therefore 
take  into  consideration  a  number  of  primary  factors  that  con- 
trol this.  These  are  as  follows: 

1.  For  larger  units: 

(a)  Decreased  unit  first  cost  of  the  equipment  with  in- 

creased1 size. 

(b)  Increased  efficiency  under  similar  conditions  for  the 

large  units. 

2.  Against  larger  units: 

(a)  Decrease  in  efficiency  at  fractional  loads.  The  larger 
the  unit,  the  smaller  per  cent  of  full  load  the  unit 
will  run  at,  under  average  conditions. 

For  example,  if  the  maximum  load  on  a  certain  power  plant 
is  10,000  kw.  and  the  mean  load  is  3000  kw.,  then  if  we  have 
only  one  unit,  it  will  run  on  the  average  at  only  30%  load, 
with  corresponding  low  efficiency.  If  on  the  other  hand  we 
put  in  two  or  more  units,  we  can  shut  down  one  unit  after  an- 
other as  the  load  decreases,  maintaining  much  nearer  full  load 
conditions  on  those  units  that  are  kept  in  operation,  and  get- 
ting correspondingly  much  higher  efficiency  under  actual 
running  conditions. 


244  TYPE  AND   SIZE  OF  UNITS 

Calling  (^4)  the  attendance  cost  per  h.p.h., 

(F)    the  fuel  cost  per  h.p.h.  at  full  load 
and  (P)   the  per  cent  of  full  load  efficiency  at  (X)  per 

cent  of  full  load 
then  at  full  load,  the  operating  cost  (a)  is 

a  =  A  +  F  per  h.p.h  ......   (98) 

and  at  fractional  loads,  the  cost  of  operation  is 

-p 


(99) 


the  load  being  under  these  conditions  ( XM) . 

If  we  are  actually  carrying  some  load  as  may  be  given  by 
some  equation  as 

N  =f(M), 

then  the  operating  cost  per  hour  is 

A  +  4 


and  the  total  operating  cost  per  year  is 

876o 


/ 


and  thus  the  production  cost  per  h.p.h.  is 


and  the  total  production  cost  per  year  is 

/876o  /*876o  /  p\ 

ScdN  +    /        \A  +j5)MdN.  (101) 

143.  The  above  are  the  conditions  for  a  single  unit  or  where 
all  the  units  are  operated,  each  carrying  a  uniform  proportion 
of  the  entire  load.  But  it  certainly  is  not  good  practice,  where 
there  are  a  number  of  units,  to  operate  them  all  at  small  loads, 
when  it  would  be  possible  to  shut  one  or  more  of  them  down 
and  operate  the  balance  at  more  nearly  full  load. 

Assuming  then  that  we  have  a  number  of  units,  MI,  Mz, 
MS)  ...  in  our  plant  all  of  the  same  size,  and  that  we  operate 
them  so  that  all  which  are  in  operation  will  be  at  full  load 


SERVICE   COST 


245 


except  one,  then  for  any  load  (M)  we  will  have,  let  us  say,  (  £7) 
units  of  (Mi)  size  operating  at  full  load  and  one  at  fractional 
load.  Under  these  conditions  the  total  attendance  cost  per 
hour  is  (AM)  and  the  total  fuel  per  hour  is 


where  UMi  +  XMi  =  Mi(U  +  X)  =  M,  .    .    (102) 

and  (X)  is  the  per  cent  of  full  load  on  the  unit  that  is  run- 
ning light  and  (P)  is  a  function  of  (X),  which  is  obtained  as 
an  approximate  equation  from  the  actual  test  (or  factory 
guarantee)  curves.  Whence  the  total  operating  cost  per  hour 
for  the  load  (M)  is 


AM  +  UMiF  +  M1          =  AM  + 


17 


But  since 


M 


£7  +  X 

and  P=f(X), 

the  total  operating  cost  is 


=  M 


A  + 


(U+  X) 


(104) 


and  the  operating  cost  per  h.p.h.  (a)  is 


u 


\ 


a  =  A  +F- 


.•-..".   (105) 


(U+  X) 
So  under  these  conditions,  the  unit  production  cost  is 

1  .Y 


_  Scp 

w 


K,=^  +  A 


FlU  + 


f(X) 


(U  + 
and  the  total  annual  production  cost  is 

F 


.    .(106) 


/876o  /*87 

ScdN+   I 


M 


A+- 


(U  +  -L-} 

\      f(x)) 


246  TYPE  AND   SIZE  OF  UNITS 

If  the  size  is  already  determined,  then  (c)  is  a  constant  and 
the  first  part  of  (K^)  above  is  simply 

pc   I  S  dN. 

But  fsdN=fNdS 

since  both  give  the  area  under  a  curve  the  equation  of  which  is 


o      i 
So  also 

Whence        sdN=       N  dS  =       tf         =      *M  =  M. 


fsdN=  f* 


Bearing  this  in  mind  will  always  simplify  this  part  of  the 
calculation. 

144.  In  the  above,  the  fixed  charges  must  be  based  on  the 
total  installation,  while  the  operating  cost  depends  on  the  units 
in  operation.  In  practice  we  should  have  to  determine  the  total 
annual  operating  costs  (a)  by  a  series  of  integration  instead 
of  only  one.  If,  for  example,  we  have  (  U  +  i)  units  in  the 
plant,  we  would  integrate  from  (  U  +  i)  to  (£/)>  then  from 
U  to  (  U  —  i),  and  so  forth,  i.e.,  always  from  X  =  100%  to 
X  =  o,  or  if  the  units  will  carry  say  25%  overload  safely, 
and  run  with  better  economy  than  below  25%  load,  then  our 
integration  steps  would  have  to  be  between  X  =  125%  and 
X  =  25%.  The  reason  for  this  is  that  at  the  point  of  shut- 
down of  a  unit  a  point  of  discontinuity  is  introduced  in  our 
curve. 

It  is  evident  then  that  units  like  generators,  steam  engines, 
and  the  like,  that  are  technically  capable  of  carrying  overloads, 
should  be  built  mechanically  well  to  carry  this  overload  for  a 
reasonable  per  cent  of  their  running  time  without  giving 
trouble,  in  order  to  allow  the  operator  to  get  the  lowest  pro- 
duction costs.  Invariably  better  economy  is  obtained  at 
reasonable  per  cent  overload  than  near  zero  load.  For  ex- 
ample, a  steam  engine  will  use  about  ij%  more  steam  be- 
tween full  load  and  25%  overload,  at  J  load  it  will  use  60% 


DETERMINATION   OF   NUMBER   OF   UNITS        247 

more  steam,  and  below  that  still  more.  A  steam  turbine  will 
use  i%  more  steam  at  25%  overload  than  at  full  load,  while 
at  J  load  it  will  use  15  to  40%  more  steam,  and  so  on. 

In  the  determination  of  the  number  of  units  (of  equal  size) 
required  in  a  plant  for  best  economy,  we  assume  that  there 
are  (n)  units  and  then  find  what  the  total  cost  of  production 
is  in  terms  of  (n).  By  plotting  this  equation,  or  by  differen- 
tiating it  and  equating  it  to  zero,  we  can  determine  the  value 
of  (n),  the  number  of  units  required  for  minimum  production 
cost,  and  thus  the  size  of  each  unit.  This  is  comparatively 
simple. 

But  we  can  obtain  better  efficiency  usually  by  using  dif- 
ferent sizes  of  units  rather  than  a  number  of  units  all  of  the 
same  size.  In  such  cases,  fractional  determination  will  give 
the  desired  results.  That  is,  when  the  load  curve  consists  of 
two  or  more  distinct  parts,  one,  let  us  say,  of  short  duration 
but  a  heavy  load,  and  the  other  of  long  duration  and  com- 
paratively light  load,  we  can  divide  the  curve  into  two  parts, 
considering  each  separately.  After  we  have  determined  thus 
the  units  best  for  each  part  of  the  load,  we  can  then  make  such 
adjustments  as  will  bring  the  two  or  more  parts  into  harmony. 

145.  Example  48.  —  A  producer  gas  power  plant  carries  a 
load  in  kw.  as  given  by  the  equation 

M  =  1000  —  o.i  AT". 
The  cost  per  kw.  of  the  units  is  given  by  the  equation 

C  =  50  —  0.005  Uj 

where  U  is  the  size  of  the  unit. 

The  fuel  cost  per  kw.h.  at  full  load  varies  with  the  size  of 
the  unit  thus: 

F  =  0.007  —  0.0000006  U. 

At  fractional  loads,  the  per  cent  of  full  load  efficiency  ob- 
tained is 

P  =  0.5(1  +  X), 

where  (X)  is  the  per  cent  of  full  load. 

The  attendance  cost  is  constant  at  $0.003  Per  kw.h. 


248  TYPE  AND   SIZE  OF  UNITS 

Assuming  fixed  charges  at  10%,  how  many  units  of  the 
same  size  will  be  required  to  give  minimum  production  cost? 
Solution:   (i)  Total  fixed  charges  per  year  are 
M0cp  =  1000  X  o.io  X  (50  —  0.005  U) 

=  5000  -  0.5  U, 
where 

M 

U  =  — -,  (n)  being  the  number  of  units  employed  in  the  plant, 
n 

,  TT       1000 

whence  U  = , 

n 

,  ,,  .   1000  ^oo 

and  M0cp  =  5000  —  0.5  H =  5000  —  - —  • 

n  n 

(2)  The  total  attendance  cost  per  year  is 

A  =  0.003   I  M  dN 
where 
I  M  dN  gives  the  total  number  of  kw.h.  produced  per  year 

/8?6o 
(1000  —  o.iN)dN 

=  0.003  (loooN  -  o.o5N2)o8760 
or  A  =  0.003  X  4,923,120  =  $14,769.36. 

(3)  Fuel  Cost:  If  the  unit  runs  at  full  load,  the  fuel  cost  is 
(F)  per  kw.h.  given  above.     If  it  runs  at  x%  of  full  load,  the 

/F\ 
fuel  cost  is  |  -  l  =  f. 


(a)  Case  of  one  unit: 

T     .  .  v      M         M 

In  this  case  X  =  —  = 


M0      1000 
since       U  =  M0  =  1000  kw., 
so  that   P  =  0.5(1  +  o.ooiM) 

and         F  =  0.007  ~~  0.0000006  X  1000  =  0.007  ~~  0.0006, 
or  F  =  0.0064,    • 

and  the  actual  fuel  cost  is 

,  _  F  _  0.0064  0.1028  ,      , 

•~  P         0.5(1  +  o.ooiM)  ~  (i  +  o.ooi'AO  PC 


DETERMINATION  OF   NUMBER   OF   UNITS 

For  the  load  (M)  the  fuel  cost  per  hour  is 

TUft  _  O.OI28  M 

(i  4-  o.ooiM) 
and  the  total  annual  fuel  cost  is 

>.oi28  MdN 


249 


But  since 

and 
whence 
so  that 


+  o.ooi  M) 
M  =  1000  —  o.iNj 
1000  —  M 


,r 
N  = 

O.I 

dN  =  —  lodM, 


(  ,rv 

=  10(1000  —  M) 


/MfdN  =  -0.128  r 
J    1000     (.1 


MdM 


+  o.ooi  M) 


o.ooiM)  - 


(O.OOIj 
=   $38,400. 

Summary  of  Case  (a),  one  unit. 
The  fixed  charges  are 


^ 
5000  —  -   -  =  $4500. 

The  attendance  costs  are  $14,769.36,  and  the  fuel  cost  is 
,400,  whence  the  total  annual  production  cost  is 

4500  +  14,769.36  +  38,400  =  $57,669.36. 
(b)   Case  of  two  units: 

In  this  case  we  have  two  units  each  of  500  kw.  The  first 
unit  runs  at  fractional  load  from  C  to  B,  then  at  full  load  from 
B  to  E.  The  second  unit  cuts  in  at  B,  carrying  the  load  area 
BDE.  Or  the  units  may  carry  the  load  area  ABDO,  both 
being  equally  loaded. 

Since  M  =  1000  —  o.i^V,  when  M  =  500, 

N  =  5000  hours. 
The  fuel  cost  for  the  load  BC  is 


760  /•876o         p 

MfdN  =  M-dN. 

o  »/  5000  * 


250 


TYPE  AND   SIZE  OF   UNITS 


But  when     U  =  500, 

F  =  0.007  —  0.0000006  X  500.0  =  0.0067 

and  X  = =  0.002  M , 


whence      BC 


500 

13760  M  X  0.0067 

5000       0.5(l    +  0.002  M) 

MdM 

-  0.134 


-L 


r 

Jsoo       I 


0.002  M 


or     BC  =  - 


0.134 


(0.002)2 
$939340. 


So  also 
BD  =  -  o. 


FIG.  52.  —  Case  of  two  units. 
MdM 


+  0.002  M 


whence 
and  finally 


-  loge(l   +  0.002M) 


Jsoo 


BD  =  $10,281.15 

/Sooo 
500  X  0.0067 

1000 
=  $16,750. 


soo 


UNITS   IN  A  PLANT  FOR   GIVEN  LOAD 

Summary  of  Case  (b)  : 
The  total  fuel  cost  is 

16,750  +  10,281.15  +  9393-40  =  $36.424.55, 
and  the  total  fixed  charges  are 


251 


* 
5000  -  ^—  =  $475°- 

The  total  annual  production  cost  is  therefore 

36,424.55  +  4750  +  14,769-36  =  $55,943-91. 

(c)  Case  of  Three  Units: 

In  the  case  of  three  units,  each  is  of  333.33  kw.  size.     The 
first  unit  runs  at  fractional  load  until  it  reaches  full  load,  after 

M 


FIG.  53.  —  Case  of  three  units. 

which  it  continues  to  run  at  full  load,  while  the  excess  is  taken 
care  of  by  cutting  in  a  second  unit.  When  the  load  exceeds 
666.67,  the  third  unit  is  cut  in.  In  this  case  the  fixed  charges 


are 


5000  ^          =  $4833.33. 


And  as  before,  the  attendance  charges  are  $14,769.36.     With 
units  of  333.33  kw.  size,  the  full  load  fuel  cost  per  kw.h.  is 
F  =  0.007  ~  0.0000006  X  333-33 

=  0.0068 

and  P  =  0.5(1  +  X), 

M 


where 


X  = 


333-33 


=  0.003^. 


252  TYPE  AND   SIZE  OF  UNITS 

So  that  P  =  0.5(1  +  o.oo3M), 

and  /  =  -  =  -        °-°°68 = °-°136 

P      0.5(1  +  o.oo3M)       (i  +  0.003 

For  the  load  area  EDHJ,  the  fuel  cost  is 

ED  =    flffdN  =   f,M*°-013*    X  dN, 
J  J    (i  +  0.003^) 

ED  '    f       MdM 


-  o.i36/ 


since          M  =  1000  —  o.iN. 

We  obtain  the  limits  of  the  above  integration  for  the  value 
of  Mj  when  N  =  8760,         or        M  =  124,  whence 

zrn  *    f124        MdM  -136    f/ 

ED  =  -  0.136  /        - — —  =  -       ^        (i-f  .oojJ£) 

^333.33(1  +  -oo3M)  (.oo3)2L 

r4 
=   $3795-91, 
3-33 

when  M  =  333-33, 

N  =  6666.67  hours 
and  M  =  666.67, 

N  =  3333-33  hours- 

The  fuel  cost  during   the  load   area  £707  and  GFIK  is 
therefore  respectively 

El  =  333-33  X  6666.67  X  0.0068  =  $15,111.11 
and  FG  =  333.33  X  3333-33  X  0.0068  =  $7555-55- 

The  fuel  cost  during  the  load  area  FEK  is 

r       Mdu 

FE  =  -  0.136   /        — -    — -, 

./333. 33  (J    +  0.003<Of) 

so  that 

FE  =  -  ,°-I36T(i  +  o.oo3M)-  log  (i  +  o.oo3M)l 

(o.oo3)2|_  J333.33 

=  $4637.90. 

And  since  the  fuel  cost  during  the  load  area  CFG  is  equal  to 
that  during  the  FEK,  we  have 

CF  =  $4637.90. 


UNITS   IN   A   PLANT   FOR    GIVEN   LOAD  253 

The  total  fuel  cost  is  therefore 

3795.91  +  15,111-11  +  7555-55  +  2  X  4637-90  =  $35,738.37 
and  the  entire  production  cost  is 

$35,738.37  +  14,769-36  +  4833.33  =  $55,34i.o6. 

(d)  Case  of  Four  Units: 

In  this  event,  each  unit  is  of  250  kw.  capacity. 

The  fixed  charges  are 

5000  -  55?  =  $4875. 
4 

The  attendance  cost  is  as  before  $14,769.36. 


u  K      L 

FIG.  54.  —  Case  of  four  units. 

To  obtain  the  fuel  cost,  note  that  FD  =  1C  =  JB,  when 

EF  =  250  kw.,  N  =  7500 
when  M  =  750;  N  =  2500 

when  M  =  500;  N  =  5000. 

The  fuel  cost  for   the   load  areas  FDGI  plus  GCHJ  plus 
HBKO  is 

250(2500  +  5000  +  7500)^, 


254  TYPE  AND   SIZE  OF  UNITS 

where      F  =  0.007  —  0.0000006  X  250  =  0.00685. 
Therefore  this  fuel  cost  is 

250  X  15,000  X  0.00685  =  $25,687.50. 
So,   also,   the  load   areas    EFD  =  DIG  =  CJB.     The   fuel 


cost  during  one  of  these  load  areas  is,  since  X  = 

C^  fMXo.oo6S$dN  r 

I  MfdN  =  I  —  -P—     -^-Tfi  =  -  0.137   /  r 
J  J  0.5(1  +  0.004^0  J  (i 

_  °-T37  r 
(0.004)  2L 

=  $2627.83. 


M_  = 
250  ~ 
MdM 


O.OO4M) 

]o 
250 


And  for  the  three  triangular  areas,  the  fuel  cost  is 
3  X  2627.83  =  $7883.49. 

Finally,  for  the  load  area  BALK,  the  fuel  cost  is 


BA  =  - 


o.oo4M)  - 


o. 004^0 

J250 


(o.oo4) 
=  $1829.81. 

Whence  the  total  fuel  cost  is 

25,687.50  +  7883.49  +  1829.81  =  $35,400.80, 

and  the  total  annual  production  cost  is 

35,400.80  +  14,769.36  +  4875  =  $57,045.16. 

( 

TABLE  93 
GENERAL  SUMMARY 


Number  of  units 

Production  cost 

I 

$57,669.36 

2 

3 
4 

55,943-91 
55,34l-o6 
57,045.16 

It  is  evident  then  that  under  the  conditions  assumed  in  the 
example,  three  units  will  give  the  best  economy,  while  to  have 


DESIGN  OF  PLANT   FOR   BEST   ECONOMY        255 

four  units  —  only  one  too  many  —  gives  nearly  as  poor  economy 
as  having  only  one. 

As  compared  with  one  unit,  the  three  units  save  in  total 
production  cost 

57,669.36  -  55,341.06  =  $2,328.30 

which  capitalized  at  5  %  equals  the  very  neat  sum  of  $46,566. 
Under  the  conditions  assumed,  then,  a  three-unit  plant  would 
be  worth  $46,566  more  than  the  same  plant  with  only  one 
unit.  And  there  would  be  very  nearly  this  same  difference  in 
value  between  a  four-  and  three-unit  plant.  But  under  the 
conditions  assumed  in  this  example,  the  entire  plant  under 
the  worst  conditions  (  U  =  o)  would  have  a  first  cost  of  only 
$50,000.  Therefore,  a  proper  choosing  of  the  number  of  units 
has  resulted  in  practically  doubling  the  value  of  the  plant! 

As  a  further  illustration  of  the  application  of  financial  en- 
gineering to  design  of  a  system,  we  give  the  following  com- 
paratively simple  example. 

146.  Example  49.  —  It  is  desired  to  install  a  pumping  plant 
of  1000  g.p.m,  capacity  to  pump  against  a  static  head  of  50  feet. 
Length  of  pipe  line  required  is  200  feet.  Determine  the  sizes 
and  types  of  each  part  of  the  system  for  the  following  operat- 
ing conditions:  Length  of  service,  5  months  at  10  hours  per 
day,  cost  of  electric  power  if  used  2  cents  per  kw.h.,  cost  of 
#2  distillate  6  cents  per  gallon,  interest  5%. 

Solution:  i  (a)  Belted  Electric.  —  For  this  type  of  plant  we 
will  have: 

Cost  of  6"  belted  centrifugal  pump $150.00 

Cost  of  25  h.p.  3  phase,  1800  r.p.m.  motor  with  rails 

and  pulley 390.00 

Cost  of  belt 21.00 

Cost  of  building  size  14'  X  22' 345-oo 

Cost  of  foundations 35-oo 


Total $941. 


oo 


The  power  consumed  based  on  a  pump  efficiency  of  60% 
and  belt  efficiency  of  85%  will  be  24.7  h.p.  at  the  motor. 


256  TYPE  AND   SIZE  OF  UNITS 

i  (b)   Direct-connected  Electric.  —  In  this  case,  we  will  have: 

Cost  of  6"  centrifugal  pump  with  sub-base  and  flexible 

coupling  for  direct  connection $300.00 

20  h.p.  3  phase,  220  volt,  1200  r.p.m.  motor  less  rails 

and  pulley 410.00 

Foundations ,.  20.00 

Cost  of  building  n'  X  14' 190.00 

Total $920.00 

The  power  consumed  at  the  motor  is  guaranteed  at  20  h.p. 

Comparison.  —  A  comparison  shows,  at  once,  that  the  cost 
of  this  part  of  the  installation  is  practically  the  same  in  either 
case,  the  reduced  size  of  building  fully  offsetting  in  the  latter 
case  the  increased  cost  of  the  direct  connected  pump.  But 
the  life  of  the  direct  connected  pump  and  motor  is  consider- 
ably greater  than  the  belted  pump  and  motor,  due  to  the 
absence  of  belt  strains,  while  the  life  of  the  belt  is  always 
comparatively  short.  Besides  this  we  have  a  saving  in  the 
latter  case  of  approximately  5  h.p.  costing  io^f  per  hour  or 

o.io  X  10  X  150  days  =  $150  per  year, 

having  a  vestance  of  $3000. 

There  is  therefore  no  question  of  our  deciding  in  favor  of 
the  direct-connected  over  that  of  the  belted  outfit.  In  fact 
the  latter  would  be  rather  a  liability. 

(2)  Oil  Engine  Plant.  —  In  this  case,  we  will  have: 

Cost  of  6"  belted  centrifugal  pump $150.00 

"     "    25  h.p.  semi-Diesel  oil  engine.  .  .  .  1200.00 

"     "    belt 22.00 

"     "   building  16'  X  28' .  .  . 500.00 

"     "   foundations.  180.00 


Total  cost $2052.00 

Cost  of  Operation  —  Engine  Plant.  --  The  full  25  h.p.  will  be 
required.  At  o.6#  of  fuel  oil  guaranteed  per  h.p.h.,  the 
engine  will  consume  15$  or  2  gallons  of  oil  per  hour  costing  12^. 


DESIGN   OF   PLANT   FOR   BEST   ECONOMY 


257 


This  gives  a  cost  of  0.12  X  10  X  150  days  =  $180  per  year 
(5  mo.). 

Cost  of  attendance  per  year     =  $60 
"     "  oil,  waste,  and  supplies  =   50 

The  total  operating  cost  is  then 

180  -f  60  +  50  =  $290. 

Cost  of  Operation,  D.C.  Electric  Plant.  —  The  output  of  the 
motor  was  20  h.p.  With  a  guaranteed  efficiency  of  89.5%, 
this  gives  an  imput  requirement  of  22.34  h.p.,  costing  44.68 
cents  per  hour  or 

0.4468  X  10  X  150  days  =  $670.20  per  year  (5  mo.) 
Cost  of  attendance     =       33.00 
"     "oil  and  waste  =       21.60 
Total  per  year  (5  mo.)  =  $724.80 

A  comparison  shows  that  the  first  cost  of  the  engine  plant 
is  considerably  over  twice  the  first  cost  of  the  direct  connected 
electric  plant,  but  the  cost  of  operation  of  the  former  is  con- 
siderably less  than  half  that  of  the  latter.  To  determine  the 
comparative  value,  and  thus  reach  a  decision,  we  proceed  as 
follows : 

TABLE  94 
DIRECT  CONNECTED  ELECTRIC  PUMPING  PLANT 


Item 

Cost  (C) 

Life,  years 

Term  factor 

Deprec. 
vestance, 
C+  T 

Pump  
Motor  

$300.00 
410.00 

25 
25 

o  .  70469 
o  .  70469 

$426.00 
581  .OO 

Foundation  
Building          

20.00 
190.00 

Permanent 
15 

I  .  OOOOO 

0.51897 

2O.OO 
367  .  oo 

Total       depreciation 
vestance 

$1304.00 

Cost  of  operation  $724.80  per  year,  giving  an  operating  vest- 
ance of  724.80  -f-  .05  =  $14,496. 

The  total  vestance  is  then 

$14,496  +  1394  =  $15,890. 


258 


TYPE  AND   SIZE  OF  UNITS 


TABLE  95 


Oil  Engine  Plant. 


Item 

Cost  (C) 

Life,  years 

Term  factor 

Deprec. 
vestance 
C  +  T 

Pump  

$150.00 

1C 

o  <;i8o7 

$280  oo 

Engine 

I2OO   OO 

JC 

O    C.  1  80  7 

Belt 

22    OO 

c. 

O    2  C.27  C 

87  oo 

Building        

5  oo  .  oo 

1C 

O    «Cl8o7 

066  oo 

Foundation.  

iSo.OO 

Permanent 

I    OOOOO 

1  80  oo 

Total       depreciation 
vestance     .    .    . 

$3842  oo 

The  total  cost  of  operation  per  year  was  $290,  giving  an 
operating  vestance  of 

$290  •*-  .05  =  $5800. 
The  total  vestance  is  then 

3842  +  5800  =  $9642 

as  compared  with  a  total  vestance  of  $15,890  for  the  direct  con- 
nected electric  plant.  The  value  of  the  engine  plant  as  com- 
pared with  the  electric  plant  is  as  15,890  to  9642  or  1.55. 
That  is  the  engine  plant  is  worth  a  little  over  1.5  times  as 
much  as  the  electric  plant. 

Should  we  use  #i  distillate  in  the  engine  costing  8  cents  per 
gallon,  instead  of  the  #2  distillate,  our  fuel  cost  per  year  would 
increase  $60,  corresponding  to  an  increase  in  operating  vest- 
ance of  $1200. 

The  total  vestance  under  these  conditions  would  be 

9642   +   I2OO   =  $10,842, 

remaining  still  far  superior  to  the  electric  plant.  Our  deci- 
sion on  this  part  of  the  plant  must  then  be  in  favor  of  the  oil 
engine  pumping  equipment. 

Discharge  Piping.  —  The  first  problem  here  would  be  to 
decide  on  the  class  of  pipe  to  be  used,  i.e.,  standard  wrought 
iron,  cast  iron,  riveted  steel  or  wood  pipe.  We  have  been 


DESIGN  OF   PLANT  FOR   BEST   ECONOMY 


259 


present  many  times  when  the  representatives  of  the  manu- 
facturers of  each  class  have  argued  the  merits  of  their  own 
and  sometimes  the  demerits  of  other  classes.  Such  special 
pleading,  however,  gets  us  nowhere.  The  tendency  of  city 
councils  is  usually  to  buy  as  expensive  pipe  as  they  can  or 
think  they  can  afford.  But  the  question  is  merely,  which  will 
render  the  service  cheapest. 

In  this  case  we  have  a  very  low  head  to  pump  against. 
Pipe  strong  enough  to  stand  very  high  heads  is  therefore  en- 
tirely unnecessary.  The  cost  of  operation  of  pipe  is  the  cost 
of  the  friction-horse-power  consumed  by  the  pipe.  This  is 
lowest  for  wood  pipe.  Under  such  conditions,  we  should  de- 
cide on  planed  machine-banded  wood  pipe. 

We  must  now  determine  the  proper  size  of  this  pipe  as 
follows : 

We  have  found  that  the  cost  of  operation  per  year  (150 
days  @  10  hrs.)  was  $290  total.  The  cost  per  hour  is  then 

290  -T-  1500  =  $0.1933, 
and  the  cost  per  h.p.h.  is 

0.1933  -h  25  =  $0.00773, 
whence  we  determine  as  follows: 


TABLE  96 
MACHINE-BANDED  WOOD  PIPE  FOR  1000  G.P.M.  AND  50"  HEAD 


Size 
in.  dia. 

Cost  (C) 
per  100' 

Life, 
years 

Depre- 
ciation 
vestance 

Friction 
head,  ft. 
per  100' 

Friction 
h.p. 

Operating 
vestance 

Total 
vestance 
per  100' 

6" 

$20.50 

25 

2Q.IO 

10.3 

5-15 

$1198.00 

$1227.10 

8" 

25-25 

25 

35-80 

2.51 

1-255 

292.00 

327.80 

10* 

33-25 

25 

47-20 

0.83 

0-4I5 

96.  2O 

143.40 

12* 

37-75 

25 

53-50 

0-34 

0.170 

39-50 

93.00 

14" 

48.00 

25 

68.10 

0.16 

0.080 

18.56 

86.66 

16" 

62.00 

25 

88.00 

0.09 

0.045 

10.44 

98.44 

An  inspection  shows  that  the  total  vestance  is  a  minimum  for 
14"  pipe,  this  size  giving  best  financial  efficiency.  For  1000 
g.p.m.  and  prices,  etc.,  as  listed,  this  size  is  best,  irrespective  of 


260 


TYPE  AND   SIZE  OF  UNITS 


the  length  of  the  line.  For  more  expensive  pipe,  a  smaller 
size  would  give  best  financial  efficiency. 

It  has  been  customary  in  an  installation  of  this  type  to  use 
eight  inch  pipe  as  best.  But  it  is  evident  that  with  this  size  of 
pipe,  the  conveyance  of  the  water  would  cost  us  nearly  four 
times  as  much  as  with  fourteen  inch  pipe. 

The  effect  of  increased  power  cost  is  to  force  us  to  use  larger 
pipe,  while  decreased  power  cost  permits  us  to  use  smaller 
pipe  for  best  financial  efficiency.  It  is  just  as  great  an  error 
to  use  too  expensive  equipment  as  too  cheap. 

We  have  now  completed  all  of  the  plant  except  intake  canal, 
screen  and  trash  racks,  suction  pipe,  check  valve  and  hand 
primer,  together  with  necessary  increasers,  bolts,  gaskets, 
flanges,  etc. 

These  items,  exclusive  of  the  check  valve,  amount  to  $187.50. 

The  check  valve  offers  a  special  problem.  If  a  six  inch  swing 
check  is  used,  it  will  introduce  a  friction  head  of  5.1  feet  cost- 
ing $12  per  month  for  power.  This  amount  is  enormous, 
being  over  10%  of  the  total  power  required.  The  use  of  a 
larger  check  valve  will  help  some  as  shown  in  the  table  below. 

TABLE  97 
SWING  CHECK  VALVE,  1000  G.P.M.  CAPACITY 


Size 
in.  dia. 

Cost  (C) 

Life, 
years 

Depre- 
ciation 
vestance 

Friction 
head 

Friction 
h.p. 

Operating 
vestance 

Total 
vestance 

6" 

$18.90 

25 

26.90 

5-1 

2.65 

$615.00 

$641.90 

8" 

3I-50 

25 

44.90 

1-25 

0.625 

145  .  oo 

189.90 

10" 

42.00 

25 

59.60 

0.41 

0-255 

59.20 

118.80 

12" 

63.00 

25 

89.50 

O.iy 

0.085 

19.80 

109.30 

14" 

84.00 

25 

119.60 

0.08 

0.04 

9-30 

128.90 

This  analysis  shows  that  the  twelve  inch  check  valve  is  best. 
But  if  this  is  so,  then  the  discharge  nozzle  of  a  1000  g.p.m.  pump 
should  be  twelve  inch  instead  of  six  inch,  with  a  corresponding  in- 
crease in  the  size  of  the  diffuser  and  suction.  The  better  thing  to 
do  would  be  to  use  a  gate  valve  in  which  the  friction  is  practically 


DESIGN  OF  PLANT  FOR  BEST  ECONOMY 


261 


zero,  when  the  gate  valve  is  wide  open  as  it  will  always  be 
when  the  pump  is  in  operation.  The  gate  valve  will  serve 
perfectly  well  for  priming  as  well  as  the  check  valve.  In  case 
of  stoppage  of  the  engine,  there  will  be  nothing  to  prevent  a 
reversal  of  the  flow  in  the  pipe  line.  If  the  end  of  the  dis- 
charge is  above  the  water  line  in  the  reservoir  or  canal,  this 
reversed  flow  can  be  limited  to  the  water  in  the  pipe  line,  and 
thus  amount  to  very  little.  We  should,  in  this  installation, 
choose  a  six  inch  gate  valve  placed  directly  on  the  nozzle  of  the 
pump,  costing  $24.50,  and  having  a  total  vestance  of  $34.40 
instead  of  $109.30  for  the  best  check  valve. 


TABLE  98 
SUMMARY  OF  PLANT  ITEMS  (ALL  COSTS  INSTALLED) 


Item 

Size 

First  cost 

Depre- 
ciation 
vestance 

H.p. 

Operating 
cost  per 
year 

Vestance 

Total 
vestance 

$5l6l.OO 
2320.00 
957.00 
966.00 
180.00 
154.80 

34-40* 
21  I.  2O 

$9984.40 

Pump 

6" 

25  h-p. 
8" 
16'  X  28' 
12  yds. 
1  4"  X  200' 
6* 

$150.00 
I  2OO.OO 
22.OO 
500.00 
iSo.OO 
96.00 
24.50 
187.50 

$289.00 
2320.00 
87.00 
966.00 
180.00 
136.20 
34-40 
211.  2O 

21 

$243  .  60 

$4872.00 

Engine  

Belt  

3-75 

43-50 

870.00 

Building  
Foundation  .  .  . 
Pining    . 

18.60 

0.08 

0-93 

Gate  valve.  .  .  . 
Incidentals.  .  .  . 

Totals  

$2360.00 

$4223.80 

$24.83 

$288.03 

$5750.60 

There  is  still  one  item  open  to  discussion  in  the  above  and 
that  is  the  belt,  this  little  item  of  $22  costing  $957  in  vestance. 
The  only  substitute  we  could  here  use  would  be  a  silent  chain 
drive  or  reduction  gear  running  in  oil,  together  with  a  common 
sub-base  for  engine  and  pump.  The  efficiency  of  such  transmis- 
sion is  guaranteed  at  98%  and  will  cost  $160  for  the  base  and 
$120  for  the  transmission.  Under  these  conditions  we  would 
have  the  following: 


262 


TYPE  AND   SIZE  OF  UNITS 


TABLE  99 


Item 

First  cost 

Life,  years 

Depre- 
ciation 
vestance 

Operating 
cost  per 
year 

Operating 
vestance 

Total 
vestance 

Base  

$160.00 

Permanent 

$160.00 

$160  oo 

Transmission  .... 

1  20.00 

15 

232.00 

$3.58 

$71.60 

303  .  60 

Total 

$280  oo 

$3Q2    OO 

$7   eg 

$71   60 

$A6  3    60 

But  with  such  an  outfit  we  could  reduce  the  size  of  the 
building  from  16'  X  28'  to  14'  X  18'  costing  $285,  with  a 
depreciation  vestance  of  $440.  This  represents  a  saving  of 
vestance  on  the  smaller  building  required  of 

966  —  440  =  $526 

more  than  offsetting  the  vestance  of  the  direct  geared  trans- 
mission and  sub-base.  We  would  therefore  decide  in  favor  of 
this  drive.  There  is  also  a  saving  of  two  yards  in  foundation 
to  be  taken  into  account  on  this  design.  Under  these  condi- 
tions our  plant  items,  costs,  and  vestances  would  be  as  follows : 

TABLE   ioo 
FINAL  SUMMARY  OF  PLANT 


Item 

Size 

First  cost 

Depre- 
ciation 
vestance 

H.p 

used 

Operating 
cost  per 
year 

Operating 
vestance 

$4872.00 

Total 

vestance 

Pump 

6" 
25  h.p. 

$150.00 
I  2OO  .  OO 
l6o.OO 
I  2O.OO 
.     285.00 
150.00 
96.00 
24.50 
187.50 

$289.00 
2320.00 
160.00 
232.00 
440.00 
150.00 
136.20 
34-40 
211  .20 

21 

$243  .  60 

$5l6l.OO 
2320.00 
l6o.OO 
303.00 
440.00 
150.00 
154.80 
34-40 
211  .20 

Engine  

Base 

Transmission 

0.05 

3.58 

71.60 

Building  
Foundation  .  .  . 
Piping 

14'  X  18' 
ic  yds. 

I4"  X200' 

6" 

0.08 

0-93 

18.60 

Gate  valve  .  .  . 

Totals  

$2373-00 

$3972.80 

21.13 

$248.11 

$4962  .  20 

$8935.00 

We  have  by  the  substitution  of  the  direct  geared  drive  above 
reduced  the  total  vestance  by  over  $1000  and  by  so  doing  re- 
duced the  cost  of  the  service  more  than  10%.  The  above 
plant,  the  product  of  scientific  cost  analysis  as  applied  to  this 


DESIGN  OF  PLANT  FOR   BEST   ECONOMY         263 

simple  design,  has  a  total  vestance  less  than  one-third  as  great 
as  the  average  best  plants  installed,  i.e.,  the  total  service  cost 
of  the  plant  is  less  than  one-third  of  that  found  in  the  best 
plants  of  this  size,  and  possibly  one-fifth  of  the  average  plant 
installed. 

147.  Example  50.  —  Design  a  noo  kw.  power  plant  assum- 
ing interest  5%,  land  50  cents  per  square  foot,  fuel  oil  $1.75  per 
barrel  (320$),  coal  $4  per  ton;  on  the  basis  of  18,500  B.t.u. 
per  pound  for  the  oil  and  12,000  B.t.u.  per  pound  of  coal. 
The  conditions  of  use  are  full  load  for  3000  hours  per  year. 

Solution:  Eleven  hundred  kilowatts  equals  approximately 
1500  h.p.  At  90%  mechanical  efficiency  and  direct  drive,  this 
would  demand  a  1700  i.h.p.  engine.  The  question  at  once 
arises  as  to  what  type  of  an  engine  we  should  use.  We  have 
for  consideration  (i)  internal  combustion  motor  and  (2)  steam 
engine  (a)  condensing  or  (b)  non-condensing  and  (3)  steam 
turbine  (a)  condensing,  (b)  non-condensing. 

(i)  Internal  Combustion  Motor.  —  Of  this  size  we  can  con- 
sider with  profit  only  the  Diesel  engine  and  the  producer  gas 
plant. 

(a)  In  the  case  of  the  Diesel  plant,  we  have 

Cost  of  Diesel  engine $120,000.00 

direct-connected  generator 16,000.00 

piping 1,000.00 

"     "    building 6,000.00 

"     "    foundations 5,000.00 

Switchboard  and  auxiliaries 3,000.00 

Total $151,000.00 

The  engines  are  guaranteed  at  0.43$  per  h.p.h.  or  two  barrels 
per  hour,  costing  $3.50.  Whence  the  cost  per  year  of  the 
fuel  is  $10,500.00. 

Cost  of  attendance  per  year $2,000.00 

"     "   maintenance  per  year 1,350.00 

"     "   oil,  waste,  etc 1,650.00 

Total $15,500.00 

Operating  cost  per  h.p.  year,  $10.33^. 


tt       a 
it       a 


264 


TYPE  AND   SIZE  OF  UNITS 


Assuming  the  life  of  the  Diesel  engine,  switchboard,  and 
auxiliaries  at  15  years,  that  of  the  generator,  building,  and 
piping  at  25  years,  we  may  tabulate  as  follows: 


TABLE    ioi 
SUMMARY  —  DIESEL  ENGINE  PLANT 


Item 

First  cost 
erected 

Depreciation 
vestance 

H.p. 

Operating 
cost  per 
year 

Operating 
vestance 

Total 
vestance 

Engines.  .  .  . 

$1  2O,OOO.  OC 

$231,230.00 

$231  230  oo 

Generator.  . 
Piping 

16,000.00 

I  OOO  OO 

22,700.00 
I.42O.OO 

1500 

$15,500.00 

$310,000.00 

332,700.00 
i  420  oo 

Building 

6,000.00 

8,  s  20.00 

8  5  20  oo 

Foundation  . 

5,000.00 

<  ,000.00 

5  ooo  oo 

Auxiliaries.  . 

3,000.00 

5,790.00 

20 

206.67 

4,133-00 

9,923.00 

Totals.  .  . 

$151,000.00 

$274,660.00 

1520 

$15,706.67 

$314,133.00 

$588,793.00 

(b)  For  the  producer  gas  plant,  we  would  have: 

Cost  of  engines  erected $75,000.00 

"     "   producers    "     10,800.00 

"     "    generator    " 16,000.00 

"     "    piping,  starter,  switchboard,  etc 7,000.00 

"     "    building 10,500.00 

"     "    foundations 6,500.00 

Total $125,800.00 

The  producer  plant  is  guaranteed  to  produce  a  br.h.p.h.  at 
the  engines  on  o.75#  of  coal  or  a  total  of  ii2S#  per  hour,  or 
5.625  tons  per  day  of  10  hours.  Allowing  a  standby  loss  of 
4%,  this  would  mean  a  total  consumption  of  5.85  tons  per  day 
or  1755  tons  per  year,  costing  $7022  per  year. 

For  operating  costs  per  year,  we  have: 

Fuel  cost $7,022.00 

Attendance 10,000.00 

Maintenance 1,800.00 

Oil,  waste,  etc 2,078.00 

Total $20,900.00 

Cost  per  h.p.  year  =  $13.93!. 


DESIGN  OF   PLANT   FOR   BEST   ECONOMY 


265 


Assuming  a  life  of  20  years  on  the  engine  and  producer, 
25  years  on  the  generator  and  building  and  15  years  on  the 
starter,  switchboard,  etc.,  we  may  summarize  as  follows: 


TABLE   102 
SUMMARY  PRODUCER  GAS  PLANT 


Item 

First  cost 

Depreciation 
vestance 

H.p. 

Operating 
cost  per 
year 

Vestance 

Total 
vestance 

Engine    and 
producer 

$85  800  oo 

$1  37  ^OO  OO 

$1  37  ^OO  OO 

Generator.  . 
Auxiliaries.. 
Building 

16,000.00 
7,000.00 
10  500  oo 

22,7OO.OO 
13,500.00 
Id.  OOO  OO 

1500 

20 

$20,900.00 
278.67 

$418,000.00 
5,573-00 

440,700.00 
19,073.00 

14  900  oo 

Foundation 

6  500  oo 

6  500  oo 

6  500  oo 

Totals.  .  . 

$125,800.00 

$195,100.00 

1520 

$21,178.67 

$423,573-°° 

$618,673.40 

A  comparison  shows  that  under  the  conditions  assumed, 
the  comparative  value  (valuance)  of  the  Diesel  Plant  is  5% 
greater.  We  should  therefore  decide  in  favor  of  the  latter  by 
a  very  small  margin. 

(c)  We  have  now  to  consider  the  steam  plant.  For  this 
type  of  plant,  we  have  costs  as  follows: 


Cost  of  boilers $13,500 

"     "   turbo-generators 20,500 

"     "    building 10,500 

"     "   foundations 6,500 

"     "    piping 6,000 

"     "   stokers 7, 500 

"     c<   ash  conveyor 1,000 

"     "    coal         "  7,500 

"     "    condensers 5, 600 

"     "    auxiliaries 2,500 

"     "    stock  and  breeching 3,400 

Total $84,500 


Life  in  Years 
25 
25 

25 

Permanent 
25 
15 
15 
i5 
25 


266  TYPE  AND   SIZE  OF  UNITS 

If  the  plant  is  guaranteed  to  produce  a  h.p.h.  on  1.5$  of 
this  coal,  then  the  consumption  per  hour  will  be  225o#,  to 
which  must  be  added  10%  for  standby  losses  and  auxiliary 
machinery,  making  a  total  of  2475$  or  1.2375  ton  Per  hour. 
This  means  a  consumption  of  3700  tons  per  year  at  $4  per  ton. 

We  have  then: 

Cost  of  fuel  per  year $14,800 

"     "    attendance 10,000 

"     "   maintenance 1,200 

"     "   oil,  waste,  etc 1,500 

Total $27,500 

Cost  per  h.p.  year  =  $18.33!. 

We  may  then  summarize  as  follows: 


TABLE  103 

SUMMARY  OF  STEAM  PLANT 


Item 

First  cost 

Depreciation 
vestance 

H.p. 

Operating 
cost  per 
year 

Vestance 

Total 
vestance 

Boilers  
Turbo-gener- 
ators   
Building  
Foundations. 
Piping  
Stokers.  .... 
Ash    convey- 
ors   

$13,500.00 

20,500.00 
10,500.00 
6,500.00 
6,000.00 
;f  ,500.00 

1,000.00 

7,500.00 
5,600.00 
2,500.00 
3,400.00 

$19,200.00 

29,200.00 
14,900.00 
6,500.00 
8,520.00 
14,500.00 

1,930.00 

14,500.00 
7,960.00 
4,820.00  ; 
6,570.00 

1500 
90 

$550,000.00 

$19,200.00 

579,200.00 
14,900.00 
6,500.00 

85,230.00 
6,570.00 

$27,500.00 

1,650.00 

33,000.00 

Coal  convey- 
ors   
Condensers.  . 
Auxiliaries..  . 
Stack,  etc.  .  . 

Totals  

$84,500.00 

$128,600.00 

1590 

$29,150.00 

$583,000.00 

$711,600.00 

Comparison  of  this  plant  with  the  Diesel  shows  that 

(a)  The  Diesel  plant  has  a  first  cost  1.79  times  as  much 

(b)  But  its  comparative  value  is  21  %  greater  than  the  steam 
plant.     . 


PROBLEMS  267 


PROBLEMS 

1.  Design  a  steam  plant  of  1000  h.p.  size,  to  operate  at  full  load  for  3000 
hours  per  year,  so  that  each  part  will  operate  at  maximum  financial 
efficiency,  assuming  interest  rate  5%,  coal  $4  per  ton,  with  a  heat  value 
of  5.4  h.p.h.  per  pound. 

2.  A  pipe  is  to  carry  io,ooo#  steam  per  hour  to  a  condensing  steam  engine. 
The  heat  costs  0.15  cent  per  h.p.h.    Taking  into  consideration  the  loss  in 
radiation  and  the  loss  of  available  heat  due  to  drop  in  pressure,  determine 
the  proper  size  of  pipe  for  maximum  financial  efficiency. 

3.  Taking  the  costs  of  pipe  as  given,  and  taking  into  consideration 
radiation  and  pressure  losses,  determine  the  pressure  of  steam  to  be  used 
for  conveying  heat  to  a  condensing  engine  for  best  financial   efficiency 
irrespective  of  other  than  transmission  considerations. 

4.  A  small  electric  generating  plant  is  to  carry  a  50  kw.  load  for  2000 
hours  per  year  and  a  25  kw.  load  for  another  1000  hours  per  year.    Design 
the  plant  for  best  financial  efficiency,  assuming  interest  rate   5%,  coal 
(5.4  h.p.h.  per  Ib.)  at  $5  per  ton,  fuel  oil  (7.3  h.p.h.  per  Ib.)  at  $1.85  per 
barrel,  distillate  (7.7  h.p.h.  per  Ib.)  at  25  cents  per  gallon. 

5.  In  problem  (4),  what  would  be  the  average  cost  per  kw.h.  produced? 
What  would  be  the  true  cost  per  kw.h.  during  each  period? 

6.  A  steam   turbo-generating  plant   carries  a  load  as  given  by  the 
equation 

M  =  1000  +  o.$N. 

Taking  into  consideration  the  variation  in  efficiency  of  turbo-generators 
with  size  and  load,  and  all  other  costs,  determine  the  number  of  units 
required  for  maximum  financial  efficiency. 

7.  In  problem  6,  if  the  plant  consisted  of  one  unit,  how  would  it  compare 
in  value  with  best  design? 

8.  It  is  desired  to  install  a  pumping  unit  to  deliver  5000  g.p.m.  during 
May,  June,  and  July  and  2000  g.p.m.  during  August  and  September,  against 
a  fixed  discharge  head  of  50  feet.    Centrifugal  pumps  electrically  driven  are 
to  be  used.    The  pumps  are  set  to  stand  10  feet  above  the  water  at  the  be- 
ginning of  the  pumping  season.    From  then  to  August  ist  the  suction  heat 
increases  uniformly  to  20  feet  and  then  remains  constant  until  the  end  of 
the  season.      The  power  costs  1.5  cents  per  kw.h.    Choose  the  type,  size 
and  number  of  units  for  best  financial  efficiency. 


INDEX 


Account,  stock,  23 

Annual  operating  costs,  53-54 

Attendance,  7,  15,  22 

—  cost  of,  centrifugal  pumps,  91 
gas  engines,  105 

generators,  99 

motors,  99 

oil  engines,  105 

steam  engines,  75 

Basic  costs,  59-62 
Basis  of  rates,  3-4,  187 
Belts,  vestance  of,  174,  175 
Boilers,  75-76 

—  price  of,  77-78 
Buildings,  cost  of,  78-80 

Canals,  cost  of,  113-114 
Capitalization  of  earnings,  n,  12 
Centrifugal  pumps,  attendance  cost  of, 

Qi 

—  efficiency  of,  89-91 

—  price  of,  81-88 

—  vestance  of,  162-170 

Change  point,  2,  140,  141,  145,  188,  189 

Charges,  fixed,  6,  7 

Charges,  replacement,  20 

.Collection,  6 

Comparative  value,  n,  29-30 

Comparison  of  power  units,  158-162 

Competition,  2,  10 

Complete  steam  installations,  cost  of,  80 

Compound  interest,  25-26 

Cost,  2,  4,  6,  14 

—  analysis,  1-2 

—  basic,  59-62 

—  determination,  45 

—  operating,  6,  7 

—  original,  u,  12 

—  price  (See  price) 


—  segregation,  i 

—  of  buildings,  78-80 

—  of  canals,  113-114 

—  of  complete  steam  installations,  80 

—  of  dams,  112 

—  of  hydro-electric  installations,  115- 
116 

—  of  tunnels,  113 

—  of  replacement,  11,12 

—  of  excavations,  114 

Dams,  cost  of,  112 
Depreciation,  7,  15,  16,  42 

—  rates,  36-40 

—  reserve,  18 

—  vestance,  46-49 

Design  for  best  economy,  5,  247-271 
Diesel  engines,  attendance  cost  of,  105 

—  efficiency  of,  106 

—  price  of,  101-102 

—  vestance  of,  150-152 

Direct  current  motors  (See  motors) 
Distribution,  6,  23 

Earnings,  capitalization  of,  n,  12 
Efficiency  of  centrifugal  pumps,  89-91 

—  Diesel  engines,  106 

—  gas  engines,  104-105 

—  generators.  98-99 

—  motors,  96-97,  99 

—  oil  engines,  104-105 

—  steam  engines,  72-74 
Engines  (See  gas,  steam,  Diesel) 
Equation  of  load  curve,  226-228 
Equipment,  life  of,  19,  20 
Equity,  27-30 
Excavations,  114 


Factor,  term,  31,  35 

Financial  efficiency,  maximum,  224,  239 


269 


270 


INDEX 


Fixed  charges,  6,  7 

Fractional  loads,  operation  at,  224,  246 

Friction  of  pipe,  no,  112 

Gas  engines,  attendance  cost  of,  105 

—  efficiency  of,  104-105 

—  price  of,  101-103 
Generators,  attendance  cost  of,  99 

—  efficiency  of,  98-99 

—  price  of,  95-96 

—  vestance  of,  156-158 

Hydro-electric  installations,  115-116 
Heat  transmission,  232 

Inadequacy,  17,  1 8 
Individual  systems,  8 
Induction  motors  (See  motors) 
Installations,  complete  steam,  cost  of,  80 
Installations,  hydro-electric,  115-116 
Instantaneous  unit  cost,  215-216 
Insurance,  7,  15,  21,  54-56 
Integral  systems,  8 
Interest,  7,  15-16,  25 

—  compound,  25-26 

Law  of  supply  and  demand,  10,  u 
Life,  natural,  21 

—  operating,  21 

—  of  structures  and  equipment,  19,  20 
Load  curve,  equation  of,  226-228 

—  replot,  217,  218 

Maintenance,  7,  15,  22 

Market  value  of  outstanding  liabilities, 

II,  12 

Materials  consumed,  7,  15,  21,  22 
Maximum  efficiency,  design  for,  247-271 
Minimum  costs,  224 
Motors,  attendance  cost  of,  99 

—  efficiency  of,  96,  97,  99 

—  price  of.  92-95 

—  vestance  of,  153-156,  186 

Natural  life,  21 

Number  of  units  in  a  system,  243-247 

Obsolescence,  17 

Oil  engines,  attendance  cost  of,  105 


—  efficiency  of,  104-105 

—  price  of,  100 

—  vestance  of,  146-150 
Operating  costs,  6,  7,  15 

—  annual,  53-54 

—  life,  21 

—  of  equipment  (See  efficiency,  attend- 
ance) 

—  variable,  222-223 
-  vestance,  35 

Original  cost,  n,  12 

Pipe,  price  of  standard,  107 

casing,  108 

riveted  steel,  in 

wood,  109,  in 

—  friction  of,  no,  112 

—  vestance  of  standard,  176-181 
wood,  182-183 

Power,  7,  15 

—  units,  comparison  of,  158-162 
Present  worth  of  a  depreciating  equip- 
ment, 43-44 

Price,  2,  4,  13-14 
Price  of  boilers,  77-78 

—  centrifugal  pumps,  81-88 
—  Diesel  engines,  101-102 

—  gas  engines,  101-103 

—  generators,  95-96 

—  motors,  92-95 

—  oil  engines,  100 

—  producer  gas  engines,  101-103 

—  steam  engines,  62-63 

—  transformers,  96 

—  turbo-generators,  64 
Prices,  table  of,  118-123 
Principal,  26 

Profits,  2,  13-14 

Production,  6 

Producer  gas  engines,  prices  of,  101-103 

Public  utilities,  4,  7-10 

Pumps  (See  centrifugal) 

Rate,  2,  4 

Rates,  basis  of,  2-4 

—  depreciation,  36-40 

—  of  electric  power  for  Oregon,  184-185 

—  of  interest,  16 


INDEX 


271 


Records,  60 

Rents,  7,  15 

Repair,  7,  15 

Replacement  costs,  n,  12,  20 

Replot,  217,  218 

Reserve,  depreciation,  18 

Reserve  units,  242 

Sale,  23 

Segregation,  cost,  i 

Service  modulus,  213-214 

Sinking  fund,  18 

Size  of  system,  232-240 

Size  of  units,  244-247 

Stand-by  units,  242-243 

Steam  engines,  attendance  cost  of,  75 

—  complete  installations,  cost  of,  80 

—  efficiency  of,  72-74 

—  price  of,  62-63 

—  steam  consumption,  65-74 

—  vestance  of,  134-144 
Stock  account,  23 
Storage,  33 

Structures,  life  of,  19-20 
Supply  and  demand,  law  of,  10-11 
Systems,  8 

Table  of  depreciation  rates,  40 

—  prices,  118-123 

—  term  factors,  32 
Taxes,  7,  15,  21,  54,  55 
Term  factor,  31-35 

Thermal  efficiency  (See  efficiency) 
Time  element  in  vestance,  127,  128 
Total  vestance,  50-52 
Transformers,  price  of,  96 


Transmission,  6 
Transportation,  6 
Tunnels,  cost  of,  113-114 
Turbo-generators,  price  of,  64 

Unit  cost,  one  period,  constant  load, 

187-190 
variable  load,  200 

—  two  periods,  constant  load,  200-209 
variable  load,  209-210 

—  three  periods  — ,  210-212 
,  212-213 

—  continuously  varying  load,  215-216 
Units,  stand-by,  242-243 

—  number  of,  243-247 
Uselessness,  18 
Utilities,  public,  9,  10 

Valuance,  128 
Value,  3 

—  comparative,  29-30 
Variable  operating  costs,  222-223 
Vestance,  45,  126,  129-134 

—  depreciating,  46-49 

—  of  belts,  174-175 

—  of  centrifugal  pumps,  162-170 

—  of  Diesel  engines,  150-152 

—  of  generators,  156-158 

—  of  motors,  153-156,  186 

—  of  oil  engines,  146-150 

—  of  standard  pipe,  176-181 

—  of  steam  engines,  134-144 

—  of  wood  pipe,  182-183 

—  operating,  35,  49 

—  time  element  in,  127-128 

—  total,  50-52 


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